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 = T_{1} = 370
^{o}C

Initial temperature of water = T_{2} = 20
^{o}C

Boiling point of water = T_{3} = 100 ^{o}C

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_{i}C_{i}(T_{1} - T) =
m_{w}C_{w}(T_{3} - T_{2}) +
m_{w}L + m_{w}C_{s}(T - T_{3})

(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 ^{o}C

The equilibrium temperature of the system = 176.98
^{o}C

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