A 5-kg piece of lead (specific heat 0.03 cal/g. $\degree$C) having a temperature of 80 $\degree$C is added to 500 g of water having a temperature of 20 $\degree$C. What is the final equilibrium temperature (in $\degree$C) of the system?

Question

A 5-kg piece of lead (specific heat 0.03 cal/g. $\degree$C) having a temperature of 80 $\degree$C is added to 500 g of water having a temperature of 20 $\degree$C. What is the final equilibrium temperature (in  $\degree$C) of the system?

Quizplus · Accepted Answer

The heat lost by the lead equals the heat gained by the water. Using the formula $m \cdot c \cdot \Delta T$, we set up the equation: $5000 \cdot 0.03 \cdot (80 - T) = 500 \cdot 1 \cdot (T - 20)$. Solving this gives $T = 34\degree C$.

A 5-Kg Piece of Lead (Specific Heat 0 °\degree° C) Having a Temperature of 80

A 5-Kg Piece of Lead (Specific Heat 0 $\degree$ C) Having a Temperature of 80