If 50 g of lead (of specific heat $\left. 0.11 \mathrm { kcal } / \mathrm { kg } \cdot \mathrm { C } ^ { \circ } \right)$ at $100 ^ { \circ } \mathrm { C }$ is put into 75 g of water (of specific heat $\left. 1.0 \mathrm { kcal } / \mathrm { kg } \cdot \mathrm { C } ^ { \circ } \right) \text { at } 0 ^ { \circ } \mathrm { C }$ What is the final temperature of the mixture?

Question

If  50 g of lead (of specific heat $\left. 0.11 \mathrm { kcal } / \mathrm { kg } \cdot \mathrm { C } ^ { \circ } \right)$ at $100 ^ { \circ } \mathrm { C }$ is put into 75 g of water (of specific heat $\left. 1.0 \mathrm { kcal } / \mathrm { kg } \cdot \mathrm { C } ^ { \circ } \right) \text { at } 0 ^ { \circ } \mathrm { C }$ What is the final temperature of the mixture?

Quizplus · Accepted Answer

The final temperature can be found by setting the heat lost by the lead equal to the heat gained by the water and solving for the final temperature.

If 50 G of Lead (Of Specific Heat 0.11kcal/kg⋅C∘)\left. 0.11 \mathrm { kcal } / \mathrm { kg } \cdot \mathrm { C } ^ { \circ } \right)0.11kcal/kg⋅C∘)

If 50 G of Lead (Of Specific Heat $\left. 0.11 \mathrm { kcal } / \mathrm { kg } \cdot \mathrm { C } ^ { \circ } \right)$