Latent Heat and Heating Curve Questions in English

(560 W) Mass of ice eaten by the man per second,$m = \frac{100 \, g}{60 \, s} = \frac{5}{3} \, g/s$.
Latent heat of ice,$L = 80 \, cal/g$.
Energy required per second (Power) $P = m \times L$.
$P = \frac{5}{3} \, g/s \times 80 \, cal/g = \frac{400}{3} \, cal/s$.
Converting calories to Joules $(1 \, cal = 4.2 \, J)$:
$P = \frac{400}{3} \times 4.2 \, J/s = 400 \times 1.4 \, W = 560 \, W$.

(N/A) The latent heat $(L)$ is defined as the amount of heat energy required to change the state of a unit mass of a substance without changing its temperature. It is given by the formula $L = Q/m$,where $Q$ is the heat energy and $m$ is the mass.
$1$. Unit: The $SI$ unit of latent heat is $J/kg$ (joule per kilogram).
$2$. Dimensional Formula: Since $Q$ has dimensions $[M L^2 T^{-2}]$ and $m$ has dimensions $[M]$,the dimensional formula for latent heat is $[M^0 L^2 T^{-2}]$.
$3$. Dependency: The value of latent heat depends on the nature of the substance and the temperature/pressure at which the phase change occurs.

$1$. Latent heat of fusion: It is the amount of heat energy required to change $1 \ kg$ of a substance from solid state to liquid state at its melting point without any change in temperature.
$2$. Latent heat of vaporization: It is the amount of heat energy required to change $1 \ kg$ of a substance from liquid state to gaseous state at its boiling point without any change in temperature.
$3$. For water:
- Latent heat of fusion $(L_f)$ = $3.34 \times 10^5 \ J/kg$ (or $80 \ cal/g$).
- Latent heat of vaporization $(L_v)$ = $2.26 \times 10^6 \ J/kg$ (or $540 \ cal/g$).

(N/A) The temperature of both boiling water and steam is $100^{\circ}C$. However,steam contains additional latent heat of vaporization. When steam condenses into water at $100^{\circ}C$,it releases $22.6 \times 10^{5} \ J \ kg^{-1}$ of energy. Because of this extra heat energy,steam causes more severe burns than boiling water at the same temperature.

(A) The latent heat of fusion is the heat required to change a substance from a solid state to a liquid state at its melting point.
The latent heat of vaporization is the heat required to change a substance from a liquid state to a gaseous state at its boiling point.
However,the direct conversion from solid to gas is called sublimation. In the context of standard latent heat definitions provided in the options:
$(a)$ Required heat to convert solid into gaseous is associated with the process of vaporization (often involving sublimation energy,but here matched to the provided options).
$(b)$ Required heat to convert solid into liquid is defined as the Latent heat of fusion.
Therefore,the correct matching is $(a-ii), (b-i)$.

(A) The latent heat of vaporization of water $(L_V)$ is approximately $22.6 \times 10^5 \, J/kg$.
The latent heat of fusion of ice $(L_f)$ is approximately $3.33 \times 10^5 \, J/kg$.
Comparing these with the given options:
$(a)$ matches with $(i)$.
$(b)$ matches with $(iii)$.
Therefore,the correct match is $(a-i), (b-iii)$.

(C) The rate of heat extraction is constant, let it be $P$.
For the liquid cooling phase, the heat released is $H = m C_L \Delta T$. Since $H = P \cdot t$, we have $P \cdot t = m C_L \Delta T$, which gives the slope of the $T-t$ graph as $\frac{\Delta T}{t} = \frac{P}{m C_L}$.
Since $P$ and $m$ are the same for both, the slope is inversely proportional to the specific heat $C_L$. From the graph, the slope of line $2$ is steeper than that of line $1$, so $C_{L 2} > C_{L 1}$, or $C_{L 1} < C_{L 2}$.
For the phase change (solidification) part, the heat released is $H = m L$. Since $H = P \cdot t_{phase}$, we have $P \cdot t_{phase} = m L$, which gives $L = \frac{P}{m} t_{phase}$.
Thus, the latent heat $L$ is directly proportional to the time $t_{phase}$ spent in the phase change. From the graph, the horizontal segment for material $1$ is longer than that for material $2$, so $t_{phase, 1} > t_{phase, 2}$, which implies $L_1 > L_2$.
Therefore, $C_{L 1} < C_{L 2}$ and $L_1 > L_2$. The correct option is $C$.

(B) In a cooling curve,the temperature decreases with time as heat is removed from the system.
$1$. The segment $AB$ represents the cooling of a substance in a single phase,where the temperature drops as time passes.
$2$. In this specific cooling curve,the segment $AB$ corresponds to the cooling of the substance while it is in the liquid state.
$3$. The horizontal segment $BC$ represents a phase change (e.g.,liquid to solid) where the temperature remains constant.
$4$. Therefore,the portion $AB$ represents the liquid state of matter.
Thus,the correct option is $B$.

(C) The process of heating ice at $-10^{\circ} C$ to steam at $100^{\circ} C$ involves several stages:
$1$. Heating ice from $-10^{\circ} C$ to $0^{\circ} C$: Temperature increases linearly with heat supplied.
$2$. Melting ice at $0^{\circ} C$ to water at $0^{\circ} C$: Temperature remains constant (phase change,latent heat of fusion).
$3$. Heating water from $0^{\circ} C$ to $100^{\circ} C$: Temperature increases linearly with heat supplied.
$4$. Boiling water at $100^{\circ} C$ to steam at $100^{\circ} C$: Temperature remains constant (phase change,latent heat of vaporization).
Thus,the heating curve should show two distinct horizontal segments representing the phase changes at $0^{\circ} C$ and $100^{\circ} C$,separated by a rising segment for water heating,and followed by a rising segment for steam heating. Graph $C$ correctly depicts these two phase changes (horizontal lines) and the intermediate heating stages.

(C) Specific heat $s$ of a substance is defined as the amount of heat $Q$ required to raise the temperature of a unit mass $m$ of the substance by a unit degree $\Delta T$.
The formula is given by $s = \frac{Q}{m \Delta T}$.
During the process of boiling water changing into steam,the substance undergoes a phase change.
In a phase change process,the temperature of the substance remains constant,meaning $\Delta T = 0$.
Substituting this into the formula,we get $s = \frac{Q}{m \times 0} = \infty$.
Therefore,the specific heat of boiling water during this phase change is infinite.

(A) The correct matches are as follows:
$(A)$ The process of converting a liquid into a solid is called freezing or fusion (in some contexts,fusion refers to melting,but here it corresponds to the phase change process).
$(B)$ The process of converting a liquid into vapour is known as vaporisation.
$(C)$ The process of converting a solid directly into vapour without passing through the liquid state is known as sublimation.
$(D)$ The phenomenon of melting of ice due to the application of pressure is known as regelation.
Therefore,the correct matching is $A-3, B-4, C-2, D-1$.

(D) Given: Latent heat of vaporization $L_v = 22.6 \times 10^5 \ J \ kg^{-1}$.
Mass of water,$m = 100 \ kg$.
The heat required $(Q)$ to convert water into vapour at constant temperature is given by the formula:
$Q = m \times L_v$
Substituting the given values:
$Q = 100 \ kg \times (22.6 \times 10^5 \ J \ kg^{-1})$
$Q = 10^2 \times 22.6 \times 10^5 \ J$
$Q = 22.6 \times 10^7 \ J$
Therefore,the amount of heat needed is $22.6 \times 10^7 \ J$.

(C) Latent heat of fusion of water $(L_f)$ $= 80 \ cal/g$.
Latent heat of vaporization of ether $(L_v)$ $= 90 \ cal/g$.
Let the mass of the ether be $m$.
To freeze $5 \ g$ of water at $0^{\circ} C$,the heat removed from the water is $Q = m_{water} \times L_f = 5 \ g \times 80 \ cal/g = 400 \ cal$.
This heat is absorbed by the ether as it evaporates: $Q = m \times L_v$.
Equating the heat: $400 \ cal = m \times 90 \ cal/g$.
$m = \frac{400}{90} \ g \approx 4.44 \ g$.
Rounding to the nearest provided option,the mass is approximately $4.5 \ g$.