Question

An ice of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat of the container varies with temperature T according to the empirical relation S=A+BT, where $A = 100cal/kgK$ and $B = 2\times{10^{ - 2}}cal/kgK$. If the final temperature of the container is 27°C, the mass of the container is? (Latent heat of fusion of water = $8 \times {10^4}cal/kg$, the specific heat of water $ = {10^3}cal/kgK$)
A. 0.495kg
B. 0.595kg
C. 0.695kg
D 0.795kg

Answer 1

Hint: The energy required for the melting of ice and heating of water is taken from the container. The heat energy is stored in the bulk of the container and hence, dependent on the mass of the container. So the mass of the container can be found from the total energy that the container transferred. Here the specific heat is dependent on temperature. So use integration to find the total energy transferred.

Complete step-by-step answer:
If we supply an energy $\Delta E$ to a system of specific heat capacity $S$ and mass $m$, the temperature of the body changes. This change in temperature and energy supplied is related as :$\Delta E = Sm\Delta T$

Here in question, the Specific heat is not constant and dependent on Temperature. $S = A + BT$ So here, we can assume a very tiny temperature range $dT$ (ie: between $T$ and $T + dT$) where the specific heat can be assumed to be a constant.
So If $dE$ amount of energy is transferred to the system at temperature $T$, then we can say,
$dE = SmdT$
Since the value of this constant is $A + BT$,
$dE = (A + BT)mdT$
and If $E$ amount of energy is transferred to the system to rise the temperature from ${T_1}$ to ${T_2}$,
$E = \int_{{T_1}}^{{T_2}} {(A + BT)mdT} $
Here, as the ice melts, we transfer The latent heat required for melting from the container into ice. Then as the temperature of Ice increases from 0 degrees to 27 degrees, we again supply energy. The total energy given to the system is : $E = m \times {L_f} + Cm\Delta T$ where. $C$ is the specific heat of the water and $m$ is its mass. $E = 0.1 \times 8\times{10^4} + 0.1 \times {10^3} \times 27 = 8000 + 2700 = 10700cal$
Now we use eqn (1) to find the Temperature rise of the container. Since $10700cal$ is the energy that the container gained from water, and this energy is what made the temperature of the container change from ${227^o}C$ (500K) to ${27^o}C$ (300K). So we can equate the two energies as:
$10700 = \int_{300}^{500} {(A + BT)mdT} $ $10700 = \left[ {AT + B\dfrac{{{T^2}}}{2}} \right]_{300}^{500}m$ $10700 = m\left( {A(500 - 300) + \dfrac{B}{2}({{500}^2} - {{300}^2})} \right)$ $10700 = m\left( {200A + \dfrac{B}{2}{{400}^2}} \right)$ $10700 = m\left( {200 \times 100 + \dfrac{{2 \times {{10}^{ - 2}}}}{2}({{400}^2})} \right)$ $10700 = m\left( {20000 + \dfrac{{2 \times {{10}^{ - 2}}}}{2}({{400}^2})} \right)$ $10700 = m\left( {20000 + 1600} \right)$
We are asked to find the mass of the container, which can be obtained readily as : $m = \dfrac{{10700}}{{21600}} = 0.4954Kg$ So the correct answer is option A.

Note: We need not do such long calculations in competitive exams. Always keep looking for simplifications that can be done. Here, the value of B is so small compared to A. Since $S = A + BT$, we can see that the variation of $S$ with temperature is very small. So if we assume the specific heat to be approximately constant, we get : $10700 = Sm\Delta T \approx Am\Delta T = 100 \times m \times (200)$ $m = \dfrac{{10700}}{{20000}} = 0.535Kg$ Now if $S$ was not constant and had higher values for higher temperatures, a lesser amount of mass would only be necessary to absorb all the heat energy. So the answer should be less than 0.535Kg checking the options gives the correct answer.