Question

A 6.20-kg steel ball at 27.7°C is dropped from a height of 21.7 m into an insulated container with 4.50 L of water at 10.1°C. If no water splashes, what is the final temperature of the water and steel? The specific heat of steel and water is 450J / (kg - K) and 4186J / (kg - K) respectively.

Answer 1

The final temperature of the water and steel mixture after the steel is dropped is determined as 7.12 ⁰C.

Conservation of energy

From the principle of conservation of energy, the energy lost by the steel ball is equal to the energy gained by the water.

P.E(steel) - Thermal(steel) = Thermal(water)

mgh - mcΔθ = mcΔθ

where;

• mass of water = density x volume = 1 kg/L x 4.5 L = 4.5 kg

(6.2)(9.8)(21.7) - (6.2)(450)(27.7 - T) = (4.5)(4186)(T - 10.1)

1,318.492 - 77,283 + 2790T = 18,837T - 190,253.7

114,289.192 = 16,047T

T = 7.12 ⁰C

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