A 50-g cube of ice,initially at 0.0°C,is dropped into 200 g of water in an 80-g aluminum container,both initially at 40°C.What is the final equilibrium temperature? (Specific heat for aluminum is 900 J/kg·°C,the specific heat of water is 4186 J/kg·°C,and L_f = 3.33 × 10⁵ J/kg. ) A) 17.6°C B) 9.5°C C) 12.1°C D) 20.6°C

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Quizplus · Accepted Answer

The heat lost by the water and the aluminum container is equal to the heat gained by the ice to melt and raise its temperature to the final equilibrium temperature.$$Q_{lost}=Q_{gained}$$$$m_{water}c_{water}(T_f-T_i)+m_{Al}c_{Al}(T_f-T_i)=m_{ice}L_f+m_{ice}c_{ice}(T_f-T_i)$$$$(0.200	ext{ kg})(4186	ext{ J/kg}\cdot\degree	ext{C})(T_f-40\degree	ext{C})+(0.080	ext{ kg})(900	ext{ J/kg}\cdot\degree	ext{C})(T_f-40\degree	ext{C})=(0.050	ext{ kg})(3.33	imes10^5	ext{ J/kg})+(0.050	ext{ kg})(2090	ext{ J/kg}\cdot\degree	ext{C})(T_f-0\degree	ext{C})$$$$837.2T_f-33488+72T_f-2880=166500+1045T_f-0$$$$1083.2T_f=192888$$$$T_f=\boxed{17.6\degree	ext{C}}$$

A 50-G Cube of Ice,initially at 0