Question

An ice cube of mass $0.1\text{kg}$ at $0^{\circ }\text{C}$ is placed in an isolated container which is at ${227}^{\circ }\text{C}$. The specific heat $S$ of the container varies with temperature $T$ according to the empirical relation $S=A+BT$, where $A=100\text{cal/kg K}$ and $B=2\times {10}^{-2}{\text{cal/kg K}}^2$. If the final temperature of the container (with the material in it) is ${27}^{\circ }\text{C}$, determine the mass of the container.
[Latent heat of fusion of water $=80{\text{cal/g}}^{\circ }\text{C}$, specific heat of water $=1{\text{cal/g}}^{\circ }\text{C}$]

• A. $0.495\text{kg}$
• B. $0.495\text{g}$
• C. $4.950\text{kg}$
• D. $4.950\text{g}$

Answer 1

The correct option is A $0.495\text{kg}$

Let $m$ be the mass of ice and $m_C$ be the mass of the container.
Initial temperature of ice $=273\text{K}$
& Initial temperature of container $=500\text{K}$
When the ice is added to the container at $T_1=500\text{K}$, heat is received by the ice due to which it melts.
Given, final temperature of the container (with the water inside it):
$T_2=300\text{K}$.

Heat received by ice is given by
$Q_1=mL+mC\Delta T$
$Q_1=0.1\times 1000\times 80+0.1\times 1000\times 1\times (300-273)$
$\Rightarrow Q_1=10700\text{cal}$
Heat given by the container is
$Q_2=-\underset{500}{\overset{300}{\int }}{m}_C(A+BT)dT=\underset{300}{\overset{500}{\int }}{m}_C(A+BT)dT$
$=m_C{[AT+\frac{B{T}^2}{2}]}_{300}^{500}$
Substituting the values of $A=100\text{cal/kg K}$ and $B=2\times {10}^{-2}{\text{cal/kg K}}^2$, we get
$Q_2=21600m_C\text{cal}$
From the principle of calorimetry,
Heat gained by ice = Heat lost by the container
i.e $Q_1=Q_2$
$\Rightarrow m_C=\frac{10700}{21600}=0.495\text{kg}$
Thus, option (a) is the correct answer.