A 200-g metal container, insulated on the outside, holds 100 g of water in thermal equilibrium at 22.00°C. A 21-g ice cube, at the melting point, is dropped into the water, and when thermal equilibrium is reached the temperature is 15.00°C. Assume there is no heat 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. The specific heat for the metal is closest to A) 3850 J/kg ∙ K. B) 2730 J/kg ∙ K. C) 4450 J/kg ∙ K. D) 4950 J/kg ∙ K. E) 5450 J/kg ∙ K.

Question

Quizplus · Accepted Answer

The heat lost by the water and the metal container is equal to the heat gained by the ice cube. 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 (100 g)* $$c_w$$ is the specific heat of water (4190 J/kg · K)* $$T_i$$ is the initial temperature of the water (22.00°C)* $$T_f$$ is the final temperature of the water (15.00°C)The heat lost by the metal container is:$$Q_m = m_m c_m (T_i - T_f)$$where:* $$m_m$$ is the mass of the metal container (200 g)* $$c_m$$ is the specific heat of the metal container (unknown)* $$T_i$$ is the initial temperature of the metal container (22.00°C)* $$T_f$$ is the final temperature of the metal container (15.00°C)The heat gained by the ice cube is:$$Q_i = m_i L_f + m_i c_i (T_f - T_i)$$where:* $$m_i$$ is the mass of the ice cube (21 g)* $$L_f$$ is the heat of fusion of ice (3.34 × 10⁵ J/kg)* $$c_i$$ is the specific heat of ice (2090 J/kg · K)* $$T_f$$ is the final temperature of the ice cube (15.00°C)* $$T_i$$ is the initial temperature of the ice cube (0°C)Setting $$Q_w + Q_m = Q_i$$, we get:$$m_w c_w (T_i - T_f) + m_m c_m (T_i - T_f) = m_i L_f + m_i c_i (T_f - T_i)$$Solving for $$c_m$$, we get:$$c_m = \frac{m_i (L_f + c_i (T_f - T_i)) - m_w c_w (T_i - T_f)}{m_m (T_i - T_f)}$$Plugging in the values, we get:$$c_m = \frac{(0.021 kg)(3.34 imes 10^5 J/kg + 2090 J/kg \cdot K (15.00°C - 0°C)) - (0.100 kg)(4190 J/kg \cdot K)(22.00°C - 15.00°C)}{(0.200 kg)(22.00°C - 15.00°C)}$$$$c_m = 3850 J/kg \cdot K$$Therefore, the specific heat for the metal is closest to 3850 J/kg · K.

A 200-G Metal Container, Insulated on the Outside, Holds 100