Question

In determining the specific heat of a new metal alloy 150 gm of substance is heated to 400 degrees C and then placed in a 200 gm calorimeter cup containing 400 gm of water at 10 degrees C. IF the final temperature of mixture is 30.5 degree C , What is the specific heat of alloy?

Answer 1

QUESTION:

In determining the specific heat of a new metal alloy 150 gm of substance is heated to 400 degree C and then placed in a 200 gm calorimeter cup containing 400 gm of water at 10 degree C. IF the final temperature of mixture is 30.5 degree C, What is the specific heat of alloy?

SOLUTION:

We assume that alloy wasn't melted, water wasn't boiled and no heat was transferred to calorimeter.

According to the energy conservation law:

$$c_{\text{alloy}} m_{\text{alloy}} (T_{\text{alloy, initial}} - T_{\text{final}}) = c_{\text{water}} m_{\text{water}} (T_{\text{final}} - T_{\text{water, initial}})$$ $$c_{\text{alloy}} = \frac{c_{\text{water}} m_{\text{water}} (T_{\text{final}} - T_{\text{water, initial}})}{m_{\text{alloy}} (T_{\text{alloy, initial}} - T_{\text{final}})} = 621.38 \frac{J}{\text{kg} \cdot \text{K}}$$

ANSWER:

$$c_{\text{alloy}} = 621.38 \frac{J}{\text{kg} \cdot \text{K}}$$