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 heat exchange with the surroundings.For water,the specific heat capacity is 4190 J/kg ∙ K and the heat of fusion is 3.34 × 10⁵ J/kg.What is the specific heat capacity of the metal of the container? A)1700 J/kg ∙ K B)970 J/kg ∙ K C)2300 J/kg ∙ K D)2800 J/kg ∙ K E)3300 J/kg ∙ K

Question

Quizplus · Accepted Answer

The heat lost by the water and the ice cube is equal to the heat gained by the metal container. The heat lost by the water is:$$Q_w = m_w c_w (T_i - T_f)$$where:* $$m_w$$ is the mass of the water* $$c_w$$ is the specific heat capacity of water* $$T_i$$ is the initial temperature of the water* $$T_f$$ is the final temperature of the waterThe heat lost by the ice cube is:$$Q_i = m_i L_f$$where:* $$m_i$$ is the mass of the ice cube* $$L_f$$ is the heat of fusion of iceThe heat gained by the metal container is:$$Q_m = m_m c_m (T_f - T_i)$$where:* $$m_m$$ is the mass of the metal container* $$c_m$$ is the specific heat capacity of the metal container* $$T_i$$ is the initial temperature of the metal container* $$T_f$$ is the final temperature of the metal containerSetting the heat lost by the water and the ice cube equal to the heat gained by the metal container, we get:$$m_w c_w (T_i - T_f) + m_i L_f = m_m c_m (T_f - T_i)$$Solving for $$c_m$$, we get:$$c_m = \frac{m_w c_w (T_i - T_f) + m_i L_f}{m_m (T_f - T_i)}$$Plugging in the given values, we get:$$c_m = \frac{(0.180 kg)(4190 J/kg \cdot K)(22.0°C - 15.0°C) + (0.024 kg)(3.34 	imes 10^5 J/kg)}{0.360 kg (15.0°C - 22.0°C)} = 1700 J/kg \cdot K$$Therefore, the specific heat capacity of the metal of the container is 1700 J/kg ∙ K.

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