Heat And Thermodynamics Ques 6
- An ice cube of mass $0.1 \mathrm{~kg}$ at $0^{\circ} \mathrm{C}$ is placed in an isolated container which is at $227^{\circ} \mathrm{C}$. The specific heat $S$ of the container varies with temperature $T$ according to the empirical relation $S=A+B T$, where $A=100 \mathrm{cal} / \mathrm{kg}-\mathrm{K}$ and $B=2 \times 10^{-2} \mathrm{cal} / \mathrm{kg}-\mathrm{K}^2$. If the final temperature of the container is $27^{\circ} \mathrm{C}$, determine the mass of the container.(Latent heat of fusion for water $=8 \times 10^4 \mathrm{cal} / \mathrm{kg}$, specific heat of water $=10^3 \mathrm{cal} / \mathrm{kg}-\mathrm{K}$ ).
(2001, 5M)
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Answer:
Correct Answer: 6.( 0.495 kg )
Solution:
Formula:
- Let $m$ be the mass of the container. Initial temperature of container,
$ T_i=(227+273)=500 \mathrm{~K} $
and final temperature of container,
$ T_f=(27+273)=300 \mathrm{~K} $
Now, heat gained by the ice cube $=$ heat lost by the container
$ \therefore(0.1)\left(8 \times 10^4\right)+(0.1)\left(10^3\right)(27)=-m \int_{500}^{300}(A+B T) d T $
or $10700=-m\left[A T+\frac{B T^2}{2}\right]_{500}^{300}$
After substituting the values of $A$ and $B$ and the proper limits, we get
$ m=0.495 \mathrm{~kg} $
BETA
