Ryan places 0.150 kg of boiling water in a thermos bottle. How many kilograms of ice at -12.0°C must Ryan add to the thermos so that the equilibrium temperature of the water is 75.0°C?

Question

Quizplus · Accepted Answer

The correct answer is A because 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 (0.150 kg)* $$c_w$$ is the specific heat of water (4190 J/kg°C)* $$T_i$$ is the initial temperature of the water (100°C)* $$T_f$$ is the final temperature of the water (75°C)The heat gained by the ice is:$$Q_i = m_i c_i (T_f - T_i)$$where:* $$m_i$$ is the mass of the ice (unknown)* $$c_i$$ is the specific heat of ice (2100 J/kg°C)* $$T_f$$ is the final temperature of the ice (75°C)* $$T_i$$ is the initial temperature of the ice (-12°C)Setting $$Q_w = Q_i$$, we get:$$m_w c_w (T_i - T_f) = m_i c_i (T_f - T_i)$$Solving for $$m_i$$, we get:$$m_i = \frac{m_w c_w (T_i - T_f)}{c_i (T_f - T_i)}$$Substituting the given values, we get:$$m_i = \frac{(0.150 	ext{ kg})(4190 	ext{ J/kg°C})(100°C - 75°C)}{(2100 	ext{ J/kg°C})(75°C - (-12°C))}$$$$m_i = 0.0233 	ext{ kg}$$

Ryan Places 0