Question

Suppose we have 200 grams of lead at a temperature of T_i = 80°C. It is placed in a container with 350 grams of water at Ti = 20°C. The two come to thermal equilibrium (same temperature). What is the expected final temperature of the mixture? Hint: the total energy exchange for both objects is zero.

Answer 1

Solution :

Given :

m_Lead = 200 g = 0.200 kg

m_Water = 350 g = 0.350 kg

T_Lead = 80^oC

T_Water = 20^oC

.

C_water = 4186 J/kg°C

C_Lead = 128 J/kg°C

.

Let the final temperature of the mixture be T.

Here, the total energy exchange for both objects is zero.

∴ m_water C_water (T - T_Water) = m_Lead C_Lead (T_Lead - T)

∴ (0.35)(4186)(T - 20) = (0.2)(128) (80 - T)

∴ (1465.1)(T - 20) = (25.6)(80 - T)

∴ (57.23)(T - 20) = (80 - T)

∴ (57.23) T - (1144.609) = 80 - T

∴ (58.23) T = 1224.609

⁂ T = 21.03 ^oC