Question

Let the final temp be ( T^{0} C ) Amount of heat added to the system in 4 ( operatorname{mins}(=240 mathrm{s}) )
( Q=100 times 240 mathrm{cal}=24000 mathrm{cal} )
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} mathrm{C} ), then ice at ( 0^{0} mathrm{C} ) to water at ( 0^{0} C^{prime} ), then water at ( 0^{0} C^{text {to } text { water at }} T^{0} C )
[
Q_{text {lac }}=(200 times 0.5 times 20)+(200 times 80)+(200 times 1 times(T-0))=17800+200 T
]
Similarly,
( Q_{text {ttod}}=Q_{A l}+Q_{text {ice}}=20(T+20)+17800+200 T=220 T+18200 )
Heat supplied is equal to heat absorbed by Al and ice.
( therefore 24000=220 T+18200 )
or, ( 220 T=5800 )
or, ( T=frac{5800}{220}=26.35^{0} C )

# W An aluminium container of mass 100 gm contains 200 gm of ice at - 20°C. Heat is added to the system at the rate of 100 cal/s. Find the temperature of the system after 4 minutes (specific heat of ice = 0.5 and L = 80 cal/gm, specific heat of Al = 0.2 cal/gm/°C) H o la 17. An iror 0°C to the ba E VOOR

Solution