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A 1.50-kg iron horseshoe initially at 650°C is dropped into a bucket containing

ID: 1477412 • Letter: A

Question

A 1.50-kg iron horseshoe initially at 650°C is dropped into a bucket containing 12.0 kg of water at 20.0°C. What is the final temperature of the water–horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away. A 1.50-kg iron horseshoe initially at 650°C is dropped into a bucket containing 12.0 kg of water at 20.0°C. What is the final temperature of the water–horseshoe system? Ignore the heat capacity of the container and assume a negligible amount of water boils away.

Explanation / Answer

mi = mass of iron = 1.50 kg

T = equilbrium temperature

Ti = temperature of iron = 650 C

Tw = Temperature of water = 20 C

mw = mass of water = 12 kg

heat lost by horshoe = heat gained by water

mi Ci (650 - T) = mw Cw (T - 20)

1.5 (450) (650 - T) = 12 (4186) (T - 20)

T = 28.4