Question

An ice cube of mass $0.1kg$ at $0^o$C is placed in an isolated container which is at ${227}^o$C. The specific heat S of the container varies with temperature T according to the empirical relation, S = A + BT, where A = 100 cal/kg K and B = 2 X ${10}^{-2}$ cal/kg $K^2$. If the final temperature of the container is ${27}^0$C, the mass of the container is :

(Latent heat of fusion of water = 8 X ${10}^4$cal/kg, specific heat of water = ${10}^3$ cal/kg K).

• A. 0.495 kg
• B. 0.595 kg
• C. 0.695 kg
• D. 0.795 kg

Answer 1

The correct option is A 0.495 kg

heat lost by container

$dQ\to dT$

$dQ=msdT$

${\int }_o^{Q}dQ={\int }_{500}^{300}m(A+BT)dT$

$\Rightarrow Q=m{[AT+\frac{B{T}^2}{2}]}_{500}^{300}$

$Q_e=-21600mcal$

Heat gained by ice.

$O^{0}ice\underset{L_f}{\overset{S}{\to }}{O}^{0}water\underset{C}{\overset{T}{\to }}{27}^{0}wa$

$Q_1=mL=0.1\times 80000=8000cal$

$Q_2=mc\Delta T=0.1\times {10}^3\times 27=2700cal$

$Q_g=Q_1+Q_2=10,700cal$

$Q_c=21600mcal\Rightarrow 21600m=10700$

$m=0.495kg$