A 360-g metal container,insulated on the outside,holds 180.0 g of water in thermal equilibrium at 22.0°C.A 24.0-g ice cube,at the melting point,is dropped into the water,and when thermal equilibrium is reached the temperature is 15.0°C.Assume there is no thermal energy exchange with the surroundings.For water,the specific heat is 4190 J/kg ∙ K and the heat of fusion is 3.34 × 10⁵ J/kg.What is the specific heat of the metal of the container?

Question

Quizplus · Accepted Answer

The heat lost by the water and the ice cube is equal to the heat gained by the metal container.$$Q_{water}+Q_{ice}=Q_{metal}$$$$m_{water}c_{water}(T_f-T_i)+m_{ice}L_f=m_{metal}c_{metal}(T_f-T_i)$$$$(0.180	ext{ kg})(4190	ext{ J/kg}\cdot	ext{K})(15.0\degree	ext{C}-22.0\degree	ext{C})+(0.0240	ext{ kg})(3.34	imes10^5	ext{ J/kg})=m_{metal}c_{metal}(15.0\degree	ext{C}-22.0\degree	ext{C})$$$$-2510	ext{ J}+8016	ext{ J}=m_{metal}c_{metal}(-7.0\degree	ext{C})$$$$m_{metal}c_{metal}=\frac{5506	ext{ J}}{7.0\degree	ext{C}}=786.6	ext{ J/kg}\cdot	ext{K}$$$$c_{metal}=\frac{786.6	ext{ J/kg}\cdot	ext{K}}{0.360	ext{ kg}}=\boxed{2185	ext{ J/kg}\cdot	ext{K}}$$

A 360-G Metal Container,insulated on the Outside,holds 180