Question

An aluminium container of mass $100gm$ contains $200gm$ of ice at $-{20}^{\circ }C$. Heat is added to the system at the rate of $100cal/s$. Find the temperature of the system after 4 minutes (specific heat of ice $+0.5$ and $L=80cal/gm$, specific heat of Al$=0.2cal/gm{/}^{\circ }C$)

• A. ${26.36}^{\circ }C$
• B. ${27.36}^{\circ }C$
• C. ${36.36}^{\circ }C$
• D. ${37.36}^{\circ }C$

Answer 1

The correct option is A ${26.36}^{\circ }C$

Let the final temp be $T^{0}C$
Amount of heat added to the system in 4 mins(=240 s): $Q=100\times 240cal=24000cal$

Heat supplied is equal to the heat absorbed by Al and ice. Heat absorbed by ice is used to convert it to ice at $0^{0}C$, then ice at $0^{0}C$ to water at $0^{0}C$, then water at $0^{0}C$ to water at $T^{0}C$
$Q_{ice}=(200\times 0.5\times 20)+(200\times 80)+(200\times 1\times (T-0\left)\right)=17800+200T$

Similarly, $Q_{Al}=m_{Al}{s}_{Al}(T-(-20\left)\right)=m_{Al}{s}_{Al}(T+20)=100\times 0.2\times (T+20)=20(T+20)$

$Q_{total}=Q_{Al}+Q_{ice}=20(T+20)+17800+200T=220T+18200$

Heat supplied is equal to heat absorbed by Al and ice.

$\therefore 24000=220T+18200$

or, $220T=5800$

or, $T=\frac{5800}{220}={26.36}^{0}C$