Heat Capacity, Specific Heat and Molar Specific Heat Questions in English

(C) The specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of a unit mass of the substance by $1 \text{ K}$.
It is an intrinsic property that depends on the nature of the substance and its phase.
Experimental observations show that the specific heat capacity of most substances, including water, varies with temperature.
Therefore, the statement that the specific heat capacity of water does not vary with temperature is false.
Thus, option $C$ is the correct answer.

(A) The heat energy supplied to the water is given by the formula $Q = m c \Delta T$,where $m$ is the mass,$c$ is the specific heat capacity,and $\Delta T$ is the change in temperature.
Given: $m = 500 \ g = 0.5 \ kg$,$c = 4184 \ J \ kg^{-1} \ K^{-1}$,$\Delta T = 90^{\circ}C - 30^{\circ}C = 60 \ K$.
Substituting the values: $Q = 0.5 \times 4184 \times 60$.
$Q = 0.5 \times 251040 = 125520 \ J$.
Since the volume change for liquid water during heating is negligible,the work done is approximately zero,and the change in internal energy $\Delta U \approx Q$.
Therefore,$\Delta U = 1.2552 \times 10^5 \ J \approx 1.25 \times 10^5 \ J$.

(A) The heat absorbed by a solid is given by the formula: $\Delta Q = m s \Delta T$, where $s$ is the specific heat capacity of the solid.
Given values are: mass $m = 2 \,kg$, heat absorbed $\Delta Q = 50 \,kJ = 50,000 \,J$, and change in temperature $\Delta T = 70^{\circ} C - 20^{\circ} C = 50^{\circ} C$.
Rearranging the formula to solve for $s$: $s = \frac{\Delta Q}{m \cdot \Delta T}$.
Substituting the values: $s = \frac{50,000 \,J}{2 \,kg \times 50^{\circ} C} = \frac{50,000}{100} \,J / kg^{\circ} C = 500 \,J / kg^{\circ} C$.
Therefore, the specific heat capacity of the solid is $500 \,J / kg^{\circ} C$.

(C) Given: Mass $m = 10 \ g = 0.01 \ kg$,velocity $v = 300 \ m/s$,specific heat $s = 150 \ J/kg \cdot ^{\circ}C$.
Kinetic energy of the bullet $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.01 \times (300)^2 = 0.005 \times 90000 = 450 \ J$.
Heat absorbed by the bullet $Q = 50\% \text{ of } KE = 0.5 \times 450 = 225 \ J$.
Using the formula $Q = ms\Delta T$,where $\Delta T$ is the change in temperature:
$225 = 0.01 \times 150 \times \Delta T$
$225 = 1.5 \times \Delta T$
$\Delta T = \frac{225}{1.5} = 150^{\circ}C$.
Thus,the increase in temperature is $150^{\circ}C$.

(C) The amount of heat $Q$ required to raise the temperature of a substance is given by $Q = \int m C \ dT$.
Given the specific heat $C = D T^{3}$.
Substituting this into the integral,we get $Q = \int_{20}^{30} m (D T^{3}) \ dT$.
Taking the constants $m$ and $D$ outside the integral: $Q = m D \int_{20}^{30} T^{3} \ dT$.
Evaluating the integral: $Q = m D \left[ \frac{T^{4}}{4} \right]_{20}^{30}$.
$Q = \frac{m D}{4} [ (30)^{4} - (20)^{4} ]$.
$Q = \frac{m D}{4} [ 810000 - 160000 ]$.
$Q = \frac{m D}{4} [ 650000 ]$.
$Q = \frac{650000}{4} m D = \frac{65}{4} \times 10^{4} m D$.

(A) The heat supplied at a constant rate is given by $\Delta H = mC \Delta \theta$.
Since the rate of heat supply $\frac{dH}{dt}$ is constant,we can write:
$\frac{dH}{dt} = mC \frac{d\theta}{dt}$.
This implies $\frac{d\theta}{dt} = \frac{1}{mC} \left( \frac{dH}{dt} \right)$.
Since $m$ and $\frac{dH}{dt}$ are constant,the slope of the $\theta-t$ graph is inversely proportional to the specific heat $C$ (i.e.,$\text{slope} \propto \frac{1}{C}$).
From the figure,the slope of line $B$ is greater than the slope of line $A$ (i.e.,$\text{slope}_B > \text{slope}_A$).
Therefore,$C_B < C_A$,which means the specific heat of $A$ is greater than that of $B$.