What is the minimum mass of ice at 0.0 °C that must be added to 1.00 kg of water to cool the water from 28.0 °C to 12.0 °C? (Heat of fusion = 333 J/g; specific heat capacities: ice = 2.06 J/g⋅K,liquid water = 4.184 J/g⋅K)

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Quizplus · Accepted Answer

The correct answer is A.The heat lost by the water is equal to the heat gained by the ice. 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 (1.00 kg)* $$c_w$$ is the specific heat capacity of water (4.184 J/g⋅K)* $$T_i$$ is the initial temperature of the water (28.0 °C)* $$T_f$$ is the final temperature of the water (12.0 °C)The heat gained by the ice is:$$Q_i = m_i (L_f + c_i (T_f - T_i))$$where:* $$m_i$$ is the mass of the ice (unknown)* $$L_f$$ is the heat of fusion of ice (333 J/g)* $$c_i$$ is the specific heat capacity of ice (2.06 J/g⋅K)* $$T_i$$ is the initial temperature of the ice (0.0 °C)* $$T_f$$ is the final temperature of the ice (12.0 °C)Setting $$Q_w = Q_i$$ and solving for $$m_i$$, we get:$$m_i = \frac{m_w c_w (T_i - T_f)}{L_f + c_i (T_f - T_i)}$$Plugging in the values, we get:$$m_i = \frac{(1.00 	ext{ kg})(4.184 	ext{ J/g⋅K})(28.0 	ext{ °C} - 12.0 	ext{ °C})}{333 	ext{ J/g} + (2.06 	ext{ J/g⋅K})(12.0 	ext{ °C} - 0.0 	ext{ °C})}$$$$m_i = 175 	ext{ g}$$Therefore, the minimum mass of ice at 0.0 °C that must be added to 1.00 kg of water to cool the water from 28.0 °C to 12.0 °C is 175 g.

What Is the Minimum Mass of Ice at 0