A 44.0-g block of ice at -15.0°C is dropped into a calorimeter (of negligible heat capacity)containing $100 \mathrm {~g}$ of water at 5.0°C. When equilibrium is reached, how much of the ice will have melted? The specific heat of ice is 2090 J/kg ? K, that of water is 4186 J/kg ? K, and the latent heat of fusion of water is 33.5 × 10⁴ J/kg. A)2.1 g B)21 g C)5.2 g D)52 g E)4.4 g

Question

Quizplus · Accepted Answer

The correct answer is A.The heat required to melt the ice is equal to the heat lost by the water. 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 of water* $T_i$ is the initial temperature of the water* $T_f$ is the final temperature of the waterThe heat required to melt the ice is:$$Q_i = m_i L_f$$where:* $m_i$ is the mass of the ice* $L_f$ is the latent heat of fusion of waterSetting these two equations equal to each other, we get:$$m_w c_w (T_i - T_f) = m_i L_f$$Solving for $m_i$, we get:$$m_i = \frac{m_w c_w (T_i - T_f)}{L_f}$$Plugging in the values given in the problem, we get:$$m_i = \frac{(0.100 \text{ kg})(4186 \text{ J/kg K})(5.0^\circ \text{C} - 0^\circ \text{C})}{33.5 \times 10^4 \text{ J/kg}} = 2.1 \text{ g}$$Therefore, the correct answer is A.

A 440-G Block of Ice at -15 100 g100 \mathrm {~g}100 g Of Water at 5

A 440-G Block of Ice at -15 $100 \mathrm {~g}$ Of Water at 5