Question

An 890-g iron block is heated to 370 ∘C and placed in an insulated container (of negligible heat capacity) containing 35.0 g of water at 20.0 ∘C.

What is the equilibrium temperature of this system? The average specific heat of iron over this temperature range is 560 J/(kg⋅K).

Answer in ∘C.

I have already tried 109 ∘C and 110 ∘C, so I don't know what I'm doing wrong. :(

Answer 1

Mass of iron block = m_i = 890 g = 0.89 kg

Mass of water = m_w = 35 g = 0.035 kg

Initial temperature of iron block = T₁ = 370 ^oC

Initial temperature of water = T₂ = 20 ^oC

Boiling point of water = T₃ = 100 ^oC

Final temperature of the mixture = T

Specific heat of iron = C_i = 560 J/(kg.K)

Specific heat of water = C_w = 4186 J/(kg.K)

Specific heat of steam = C_s = 1996 J/(kg.K)

Latent heat of vaporization of water = L = 2260000 J/kg

The heat lost by the iron block is equal to the heat gained by the water.

m_iC_i(T₁ - T) = m_wC_w(T₃ - T₂) + m_wL + m_wC_s(T - T₃)

(0.89)(560)(370 - T) = (0.035)(4186)(100 - 20) + (0.035)(2260000) + (0.035)(1996)(T - 100)

184408 - 498.4T = 11720.8 + 79100 + 69.86T - 6986

568.26T = 100573.2

T = 176.98 ^oC

The equilibrium temperature of the system = 176.98 ^oC