Heat capacity Questions in English

(C) The relationship between the heat supplied $(q)$,molar heat capacity at constant pressure $(C_p)$,and the temperature change $(\Delta T)$ is given by the formula: $q = n \cdot C_p \cdot \Delta T$
Given:
$q = 1.0 \, kJ = 1000 \, J$
$C_p = 75 \, J \, K^{-1} \, mol^{-1}$
Mass of water = $100 \, g$
Molar mass of water $(H_2O)$ = $18 \, g \, mol^{-1}$
Calculate the number of moles $(n)$:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, g}{18 \, g \, mol^{-1}} = 5.55 \, mol$
Substitute the values into the formula:
$1000 = (\frac{100}{18}) \times 75 \times \Delta T$
$1000 = 5.55 \times 75 \times \Delta T$
$1000 = 416.67 \times \Delta T$
$\Delta T = \frac{1000}{416.67} \approx 2.4 \, K$
Therefore,the increase in temperature is $2.4 \, K$.

(C) The molar mass of water $(H_2O)$ is $18 \ g \ mol^{-1}$.
Given,molar heat capacity $(C_p)$ = $75 \ J \ K^{-1} \ mol^{-1}$.
Specific heat capacity $(c)$ = $\frac{C_p}{\text{Molar mass}} = \frac{75}{18} \approx 4.17 \ J \ g^{-1} \ K^{-1}$.
Heat supplied $(Q)$ = $1.0 \ kJ = 1000 \ J$.
Mass of water $(m)$ = $100 \ g$.
Using the formula $Q = m \cdot c \cdot \Delta T$:
$1000 = 100 \times 4.17 \times \Delta T$.
$\Delta T = \frac{1000}{417} \approx 2.4 \ K$.

(B) The amount of heat required to raise the temperature of $1 \ mol$ of a substance by $1^{\circ}C$ (or $1 \ K$) is defined as the molar heat capacity of that substance.
Therefore,the correct option is $(B)$.

(B) The heat required to raise the temperature of a body by $1 \, K$ is defined as the thermal capacity of the body.

(B) Using Kirchhoff's equation: $\Delta H_2 - \Delta H_1 = \Delta C_p \times (T_2 - T_1)$.
Given $\Delta C_p = 7.63 \, cal/deg = 7.63 \times 10^{-3} \, Kcal/deg$.
$T_1 = 25 + 273 = 298 \, K$.
$T_2 = 100 + 273 = 373 \, K$.
$\Delta H_{100\,^{\circ}C} = \Delta H_{25\,^{\circ}C} + \Delta C_p \times (T_2 - T_1)$.
$\Delta H_{100\,^{\circ}C} = 68.3 + 7.63 \times 10^{-3} \times (373 - 298)$.

(B) The molar heat capacity $C$ is defined as $C = \frac{Q}{n \Delta T}$.
For a phase transition (like ice in equilibrium with water) at constant pressure,the temperature remains constant,meaning $\Delta T = 0$.
Since the denominator $\Delta T$ is $0$,the value of $C$ becomes $\frac{Q}{0} = \infty$.
Therefore,the molar heat capacity is infinite.

(B) The molar heat capacity at constant pressure is defined as $C_P = (\frac{\delta H}{\delta T})_P$.
At phase equilibrium between ice and water,the temperature remains constant even when heat is added,meaning $\delta T = 0$.
Therefore,$C_P = \frac{\delta H}{0} = \infty$.

(B) First,calculate the number of moles $(n)$ using the ideal gas conditions at $STP$:
$n = \frac{4.48 \ L}{22.4 \ L \ mol^{-1}} = 0.2 \ mol$.
Next,calculate the molar heat capacity at constant volume $(C_v)$:
$C_v = \frac{q_v}{n \times \Delta T} = \frac{12 \ cal}{0.2 \ mol \times 15 \ K} = \frac{12}{3} = 4 \ cal \ mol^{-1} K^{-1}$.
Finally,use the relation $C_p - C_v = R$ to find $C_p$:
Given $R \approx 2 \ cal \ mol^{-1} K^{-1}$ (as per the provided options and context),
$C_p = C_v + R = 4 + 2 = 6 \ cal \ mol^{-1} K^{-1}$.

(B) According to Kirchhoff's law,the variation of enthalpy of reaction with temperature is given by: $\Delta_r H_2 - \Delta_r H_1 = \Delta_r C_p (T_2 - T_1)$.
First,calculate $\Delta_r C_p$ for the reaction $A_{(g)} \to 2B_{(g)}$:
$\Delta_r C_p = \sum \nu_p C_{p,m}(\text{products}) - \sum \nu_r C_{p,m}(\text{reactants})$
$\Delta_r C_p = [2 \times C_{p,m}(B)] - [1 \times C_{p,m}(A)]$
$\Delta_r C_p = [2 \times 5 \ J \ K^{-1} \ mol^{-1}] - [20 \ J \ K^{-1} \ mol^{-1}] = 10 - 20 = -10 \ J \ K^{-1} \ mol^{-1}$.
Since $\Delta_r C_p < 0$,the term $\Delta_r C_p (T_2 - T_1)$ will be negative if $T_2 > T_1$ (increasing temperature).
Therefore,$\Delta_r H_2 = \Delta_r H_1 + \Delta_r C_p (T_2 - T_1)$ will be less than $\Delta_r H_1$.
Thus,$\Delta H$ will decrease on increasing temperature.

(B) For an ideal gas,the relationship between molar heat capacities is given by Mayer's relation: $C_P - C_V = R$.
At constant volume,the work done by the gas is zero $(w = 0)$ because there is no change in volume $(dV = 0)$.
Therefore,all the heat supplied at constant volume is used to increase the internal energy of the system $(q_V = \Delta U)$.
At constant pressure,some heat is used to increase the internal energy and some is used to perform work of expansion $(q_P = \Delta U + P\Delta V)$.
Since the internal energy change $(\Delta U)$ depends only on temperature for an ideal gas,it is the same for both processes for the same temperature change.
Thus,$C_P$ must be greater than $C_V$ because additional heat is required to perform work at constant pressure.

(D) For an ideal gas,the molar heat capacity at constant pressure $(C_p)$ is defined as $C_p = C_v + R$.
For a monatomic ideal gas,the molar heat capacity at constant volume $(C_v)$ is $\frac{3}{2}R$.
Therefore,$C_p = \frac{3}{2}R + R = \frac{5}{2}R = 2.5R$.
Since $C_p$ for an ideal gas is independent of temperature,the graph of $C_p$ versus $T$ is a horizontal straight line at the value of $2.5R$.

(B) The molar heat capacity is defined as $C = \frac{dq}{dT}$.
For a phase transition occurring at constant temperature and pressure (equilibrium between ice and water),the temperature change $\Delta T$ is $0$.
Since the heat added is used for the phase change (latent heat) without changing the temperature,$dT = 0$.
Therefore,$C = \frac{dq}{0} = \infty$ (infinity).

(B) Using Kirchhoff's equation: $\Delta H_2 = \Delta H_1 + \Delta C_p(T_2 - T_1)$.
Given: $\Delta H_1 = -285.8 \ kJ \ mol^{-1} = -285800 \ J \ mol^{-1}$,$\Delta C_p = 32 \ J \ K^{-1}$,$T_1 = 27 + 273 = 300 \ K$,$T_2 = 127 + 273 = 400 \ K$.
$\Delta H_2 = -285800 + 32(400 - 300)$.
$\Delta H_2 = -285800 + 32(100) = -285800 + 3200 = -282600 \ J \ mol^{-1}$.
Converting to $kJ \ mol^{-1}$: $\Delta H_2 = -282.6 \ kJ \ mol^{-1}$.

(D) The term ${\left[ {\partial U/\partial T} \right]_V}$ represents the molar heat capacity at constant volume,denoted as $C_V$.
For an ideal gas,the internal energy $U$ depends only on temperature.
For a monoatomic gas like $He$,the degrees of freedom $f = 3$.
The molar heat capacity at constant volume is given by the formula $C_V = (f/2)R$.
Substituting $f = 3$,we get $C_V = (3/2)R$.

(A) The formula for specific heat capacity $(c)$ is given by $q = m \times c \times \Delta T$.
Here,$q = 192 \ J$,$m = 45.0 \ g$,and $\Delta T = 6 \ ^oC$.
Rearranging for $c$: $c = \frac{q}{m \times \Delta T}$.
Substituting the values: $c = \frac{192}{45.0 \times 6} = \frac{192}{270}$.
$c \approx 0.711 \ J \ g^{-1} \ ^oC^{-1}$.

(D) For an ideal gas,the change in internal energy is given by the formula: $\Delta U = n C_{V} \Delta T$
Given values are:
$n = 4 \; mol$
$\Delta T = T_{2} - T_{1} = 500 \; K - 300 \; K = 200 \; K$
$\Delta U = 5000 \; J$
Substituting these values into the equation:
$5000 = 4 \times C_{V} \times 200$
$5000 = 800 \times C_{V}$
$C_{V} = \frac{5000}{800} = 6.25 \; J \; mol^{-1} \; K^{-1}$

The heat required $(q)$ is given by the formula: $q = n \cdot C_m \cdot \Delta T$
First,calculate the number of moles $(n)$ of $Al$:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{60.0 \, g}{27.0 \, g \, mol^{-1}} = 2.222 \, mol$
Next,determine the change in temperature $(\Delta T)$:
$\Delta T = 55^{\circ} C - 35^{\circ} C = 20 \, K$
Now,substitute the values into the heat formula:
$q = 2.222 \, mol \times 24 \, J \, mol^{-1} \, K^{-1} \times 20 \, K$
$q = 1066.56 \, J$
Convert the heat to $kJ$:
$q = \frac{1066.56}{1000} \, kJ = 1.067 \, kJ$

(N/A) The quantity of heat required to increase the temperature of any substance by $1^{\circ}C$ is called the heat capacity $(C)$ of that substance.
$q = C \cdot \Delta T$,where $C$ is the heat capacity.
We can measure the heat supplied by monitoring the temperature rise,provided we know the heat capacity.
When $C$ is large,a given amount of heat results in only a small temperature rise. Water has a large heat capacity,meaning a lot of energy is needed to raise its temperature. $C$ is directly proportional to the amount of substance.
Molar Heat Capacity: The molar heat capacity $(C_m)$ of a substance is the heat capacity for one mole of the substance,defined as the quantity of heat needed to raise the temperature of one mole by one degree Celsius.
$C_m = C / n$
Specific Heat Capacity: The specific heat capacity $(c)$ is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius.
$q = c \times m \times \Delta T = C \Delta T$,where $m$ is the mass of the substance.

(N/A) Given: $q = 20 \ cal$,$\Delta T = 30^{\circ}C - 25^{\circ}C = 5^{\circ}C$,$m = 15 \ g$,$M = 27 \ g \ mol^{-1}$.
$1$. Heat capacity $(C)$: $C = \frac{q}{\Delta T} = \frac{20 \ cal}{5^{\circ}C} = 4.0 \ cal \ {}^{\circ}C^{-1}$.
$2$. Specific heat capacity $(c)$: $c = \frac{C}{m} = \frac{4.0 \ cal \ {}^{\circ}C^{-1}}{15 \ g} \approx 0.267 \ cal \ g^{-1} \ {}^{\circ}C^{-1}$.
$3$. Molar heat capacity $(C_m)$: $C_m = c \times M = 0.267 \ cal \ g^{-1} \ {}^{\circ}C^{-1} \times 27 \ g \ mol^{-1} \approx 7.21 \ cal \ mol^{-1} \ {}^{\circ}C^{-1}$.

(B) The molar heat capacity $(C_{p})$ is defined as the heat capacity per mole of a substance.
Specific heat $(c)$ is defined as the heat capacity per gram of a substance.
For $1 \ mol$ of water $(H_{2}O)$,the molar mass is $18 \ g \ mol^{-1}$.
The relationship is given by: $C_{p} = M \times c$,where $M$ is the molar mass.
Therefore,for water: $C_{p} = 18 \times c$.

(B) Heat capacity $(C)$ is defined as the amount of heat required to raise the temperature of a given quantity of a substance by $1 \ K$ (or $1 \ ^\circ C$).
Mathematically,it is expressed as $q = C \times \Delta T$,where $q$ is the heat absorbed and $\Delta T$ is the change in temperature.
Its $SI$ unit is $J \ K^{-1}$.

(N/A) Specific heat capacity is defined as the amount of heat required to raise the temperature of $1 \ g$ of a substance by $1 \ K$. Its unit is $J \ K^{-1} \ g^{-1}$ or $J \ g^{-1} \ K^{-1}$.
Molar heat capacity is defined as the amount of heat required to raise the temperature of $1 \ mol$ of a substance by $1 \ K$. Its unit is $J \ K^{-1} \ mol^{-1}$ or $J \ mol^{-1} \ K^{-1}$.

(A) Heat capacity is defined as the amount of heat required to raise the temperature of a given quantity of a substance by $1^{\circ}C$ (or $1 \ K$).

(N/A) The unit of specific heat capacity is $J \cdot K^{-1} \cdot g^{-1}$ (or $J \cdot g^{-1} \cdot K^{-1}$).
The unit of molar heat capacity is $J \cdot K^{-1} \cdot mol^{-1}$ (or $J \cdot mol^{-1} \cdot K^{-1}$).

(N/A) $(i)$ First law of thermodynamics.
$(ii)$ Heat capacity.
$(iii)$ $1 \ g$ (or unit mass).

(N/A) $(i)$ $J \ K^{-1}$ (Joule per Kelvin)
$(ii)$ Extensive
$(iii)$ Molar heat capacity

(B) Specific heat capacity $(c)$ is defined as the amount of heat required to raise the temperature of $1 \ g$ of a substance by $1 \ K$ (or $1 \ ^\circ C$).
It is an intensive property because it does not depend on the amount of matter present.
Examples of intensive properties include: Melting point,density,boiling point,viscosity,and refractive index.

(N/A) $(i)$ $1 \ g$
$(ii)$ $J \ K^{-1}$
$(iii)$ Extensive
$(iv)$ Molar heat capacity

(A) The ratio of molar heat capacities at constant pressure $(C_p)$ to that at constant volume $(C_V)$ is represented by the adiabatic index $\gamma = C_p / C_V$.
For a monoatomic gas,the degrees of freedom $f = 3$,so $\gamma = 1 + (2/f) = 1 + 2/3 \approx 1.67$.
For a diatomic gas,the degrees of freedom $f = 5$,so $\gamma = 1 + 2/5 = 1.40$.
For a non-linear triatomic gas,the degrees of freedom $f = 6$,so $\gamma = 1 + 2/6 \approx 1.33$.
Given that $C_p / C_V = 1.67$,the gas must be monoatomic.
Among the given options,$He$ (Helium) is a noble gas and is monoatomic,while $H_2$,$CO_2$,and $CH_4$ are polyatomic.
Therefore,the correct option is $A$.

(C) The heat lost by the metal is equal to the heat gained by the water in an insulated container.
Let $c$ be the specific heat of the metal.
Mass of metal $m_m = 100 \, g$,Initial temperature $T_{m,i} = 80^{\circ} C$.
Mass of water $m_w = 1000 \, g$,Initial temperature $T_{w,i} = 15^{\circ} C$.
Final equilibrium temperature $T_f = 15.69^{\circ} C$.
Specific heat of water $c_w = 4.184 \, J / g \cdot ^{\circ} C$.
Heat lost by metal $= m_m \times c \times (T_{m,i} - T_f) = 100 \times c \times (80 - 15.69) = 6431c$.
Heat gained by water $= m_w \times c_w \times (T_f - T_{w,i}) = 1000 \times 4.184 \times (15.69 - 15) = 4184 \times 0.69 = 2886.96 \, J$.
Equating the two: $6431c = 2886.96$.
$c = \frac{2886.96}{6431} \approx 0.4489 \, J / g \cdot ^{\circ} C$.
Rounding to two decimal places,we get $0.45 \, J / g \cdot ^{\circ} C$.

(B) Power of the heater $P = 60 \, W = 60 \, J \, s^{-1}$.
Total energy supplied $Q = P \times t = 60 \, J \, s^{-1} \times 100 \, s = 6000 \, J$.
Since the container is adiabatic and of constant volume,the heat supplied is equal to the heat absorbed by the gas.
$Q = C \times \Delta T$,where $C$ is the heat capacity and $\Delta T$ is the change in temperature.
$6000 \, J = C \times 5 \, K$.
$C = \frac{6000}{5} = 1200 \, J \, K^{-1}$.

(B) Extensive properties are those that depend on the amount of matter present in the system.
$Heat \ capacity$ is an extensive property because it depends on the total amount of substance.
Viscosity,density,and surface tension are intensive properties as they are independent of the amount of matter present.

(A) The formula for heat energy is $q = n \times C_p \times \Delta T$.
Given:
Mass of water $(m)$ = $180 \ g$.
Molar mass of water $(M)$ = $18 \ g \ mol^{-1}$.
Number of moles $(n)$ = $\frac{m}{M} = \frac{180}{18} = 10 \ mol$.
Change in temperature $(\Delta T)$ = $15^{\circ} C - 10^{\circ} C = 5 \ K$ (or $5^{\circ} C$).
Heat energy $(q)$ = $3765 \ J$.
Substituting the values into the formula:
$3765 = 10 \times C_p \times 5$.
$3765 = 50 \times C_p$.
$C_p = \frac{3765}{50} = 75.3 \ J \ mol^{-1} \ K^{-1}$.
Thus,the correct option is $A$.

(B) Step $1$: Calculate the number of moles of water. $n = \frac{\text{mass}}{\text{molar mass}} = \frac{180 \ g}{18 \ g \ mol^{-1}} = 10 \ mol$.
Step $2$: Calculate the change in temperature. $\Delta T = 15^{\circ}C - 10^{\circ}C = 5 \ K$.
Step $3$: Use the formula for heat energy $q = n \times C_{p} \times \Delta T$.
Step $4$: Substitute the values: $q = 10 \ mol \times 75.3 \ J \ mol^{-1} \ K^{-1} \times 5 \ K = 3765 \ J$.

(C) Given:
Weight of aluminium $(w) = 54 \ g$
Temperature difference $(\Delta T) = 60^{\circ}C - 40^{\circ}C = 20 \ K$ (or $20^{\circ}C$)
Molar heat capacity of aluminium $(C_m) = 24 \ J \ mol^{-1} \ K^{-1}$
Atomic mass of $Al$ $(M) = 27 \ g \ mol^{-1}$
Number of moles of $Al$ $(n) = \frac{w}{M} = \frac{54}{27} = 2 \ mol$
Heat required $(Q) = n \times C_m \times \Delta T$
$Q = 2 \times 24 \times 20 = 960 \ J$
Hence,option $(C)$ is the correct answer.

(D) At constant volume,the change in internal energy $\Delta U$ is given by the formula: $\Delta U = n C_v \Delta T$.
Given:
Number of moles $n = 1 \ mol$.
Molar heat capacity $C_v = 20.8 \ J \ mol^{-1} \ K^{-1}$.
Change in temperature $\Delta T = 40^{\circ} C - (-20^{\circ} C) = 60 \ K$.
Substituting the values:
$\Delta U = 1 \ mol \times 20.8 \ J \ mol^{-1} \ K^{-1} \times 60 \ K$.
$\Delta U = 1248 \ J$.

(A) The formula for heat supplied at constant pressure is $\Delta H = n C_p \Delta T$.
Given:
Heat supplied $(q)$ = $1 \ kJ = 1000 \ J$.
Mass of water $(m)$ = $100 \ g$.
Molar mass of water $(M_{H_2O})$ = $18 \ g \ mol^{-1}$.
Number of moles $(n)$ = $\frac{m}{M} = \frac{100}{18} \ mol$.
$C_p = 75 \ J \ K^{-1} \ mol^{-1}$.
Substituting the values:
$1000 = (\frac{100}{18}) \times 75 \times \Delta T$.
$\Delta T = \frac{1000 \times 18}{100 \times 75} = \frac{180}{75} = 2.4 \ K$.

(C) The molar heat capacity $(C_p)$ at constant pressure for gases is generally higher than for liquids and solids because a portion of the heat supplied is used to perform work against the external pressure during expansion.
For monatomic gases like Helium $(He)$,$C_p = \frac{5}{2}R \approx 20.8 \ J \cdot K^{-1} \cdot mol^{-1}$.
For liquids like Bromine $(Br_2)$,the molar heat capacity is approximately $75.7 \ J \cdot K^{-1} \cdot mol^{-1}$ (Note: While $C_p$ for liquids can be high,in the context of standard chemistry problems comparing states of matter,gases are often considered to have higher degrees of freedom or specific heat characteristics relative to solids).
However,based on standard thermodynamic data,the molar heat capacity of liquids like $Br_2$ is often higher than that of monatomic gases. Re-evaluating the standard trend: $C_p$ (liquid) > $C_p$ (gas) > $C_p$ (solid).
Given the specific options provided in competitive chemistry contexts,the intended order is $Helium(g) > Bromine(l) > Copper(s)$ based on the degrees of freedom and expansion work.