Suppose we put a $0.3-\mathrm{kg}$ hunk of ice at $0^{\circ} \mathrm{C}$ into an insulating thermos bottle filled with $1000 \mathrm{~g}$ of water at $20^{\circ} \mathrm{C}$. What will be the system's final temperature?

Question

Quizplus · Accepted Answer

The heat lost by the water will be 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, $c_w$ is the specific heat of water, $T_i$ is the initial temperature of the water, and $T_f$ is the final temperature of the system.The heat gained by the ice is:$$Q_i = m_i L_f + m_i c_i (T_f - T_i)$$where $m_i$ is the mass of the ice, $L_f$ is the latent heat of fusion of ice, $c_i$ is the specific heat of ice, and $T_i$ is the initial temperature of the ice.Setting $Q_w = Q_i$, we get:$$m_w c_w (T_i - T_f) = m_i L_f + m_i c_i (T_f - T_i)$$Solving for $T_f$, we get:$$T_f = \frac{m_w c_w T_i + m_i L_f + m_i c_i T_i}{m_w c_w + m_i c_i}$$Plugging in the numbers, we get:$$T_f = \frac{(1000 \mathrm{~g})(4.187 \mathrm{~J/g \degree C})(20^{\circ} \mathrm{C}) + (300 \mathrm{~g})(334 \mathrm{~J/g}) + (300 \mathrm{~g})(2.09 \mathrm{~J/g \degree C})(0^{\circ} \mathrm{C})}{(1000 \mathrm{~g})(4.187 \mathrm{~J/g \degree C}) + (300 \mathrm{~g})(2.09 \mathrm{~J/g \degree C})}$$$$T_f = 0.1^{\circ} \mathrm{C}$$Therefore, the final temperature of the system will be almost exactly $0^{\circ} \mathrm{C}$.

Suppose We Put A 0.3−kg0.3-\mathrm{kg}0.3−kg Hunk of Ice At 0∘C0^{\circ} \mathrm{C}0∘C Into an Insulating Thermos Bottle Filled With

Suppose We Put A $0.3-\mathrm{kg}$ Hunk of Ice At $0^{\circ} \mathrm{C}$ Into an Insulating Thermos Bottle Filled With