Question

A (750 ^oC) 350 g aluminum ingot is placed into an 0.65 kg steel pot with a lid. There is 320 g of water in the steel pot; both are initially at room temperature, 22^oC. What is the final equilibrium temperature if the ingot is placed in the pot and then the top is quickly sealed? If there is steam produced, how much is there? Show all work.

Answer 1

m_a = mass of aluminium = 0.35 kg

T_ai = initial temp of aluminium = 750 C

c_a = specific heat of aluminium = 900

m_s = mass of steel = 0.65 kg

T_si = initial temp of steel = 22 C

c_s = specific heat of steel = 490

m_w = mass of water = 0.32 kg

T_wi = initial temp of water = 22 C

c_w = specific heat of water = 4186

T = equilibrium Temperature

Using conservation of Heat

heat lost by aluminium = heat gained by steel + Heat gained by water

m_a c_a (T_ai - T) = m_s c_s (T - T_si) + m_w c_w (T - T_wi)

(0.35) (900) (750- T) = (0.65) (490) (T - 22) + (0.32) (4186) (T - 22)

T = 138.23