50 g of lead (c = 0.11 kcal/kg. °C) at 100°C is put into 75 g of water at 0°C. What is the final temperature of the mixture?

Question

Quizplus · Accepted Answer

The final temperature can be found by setting the heat gained by water equal to the heat lost by lead. Using the formula $Q = mc\Delta T$, where $Q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature, we can set up the equation: $m_{\text{Pb}}c_{\text{Pb}}(T_f - 100) = m_{\text{H2O}}c_{\text{H2O}}(T_f - 0)$. Solving for $T_f$ with $m_{\text{Pb}} = 0.05\, \text{kg}$, $c_{\text{Pb}} = 0.11\, \text{kcal/kg.}°C$, $m_{\text{H2O}} = 0.075\, \text{kg}$, and $c_{\text{H2O}} = 1\, \text{kcal/kg.}°C$, gives a final temperature of approximately 6.8°C.

50 G of Lead (C = 0