Thermal Equality

Classical Mechanics Level 3

A ball of aluminum (cp = 0.897 J/g°C) has a mass of 100 grams and is initially at a temperature of 150°C. This ball is quickly inserted into an insulated cup containing 100 ml of water at a temperature of 15.0°C. What will be the final, equilibrium temperature of the ball and the water?(round to nearest whole number)


The answer is 39.

Mardokay Mosazghi by Mardokay Mosazghi , 22, USA

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2 solutions

Chew-Seong Cheong
Jul 24, 2014

Let the final equilibrium temperature be θ \theta . Then:

  • Heat lost by aluminum = c a m a ( θ a − θ ) = c_{a}m_{a}(\theta_{a}-\theta) = 0.897 × 100 ( 150 − θ ) = 89.7 ( 150 − θ ) =0.897 \times 100(150-\theta)= 89.7(150-\theta)

  • Heat gain by water = c w m w ( θ w − θ ) = c_{w}m_{w}(\theta_{w}-\theta) = 4.1813 × 100 ( θ − 15 ) = 418.13 ( θ − 15 ) =4.1813\times 100(\theta-15)=418.13(\theta - 15)

At equilibrium, heat lost by aluminum = heat gained by water. Therefore, 89.7 ( 150 − θ ) = 418.13 ( θ − 15 ) 89.7(150-\theta)=418.13(\theta - 15) 507.83 θ = 19726.95 507.83\theta=19726.95 θ = 38.85 ≈ 3 9 ∘ C \theta = 38.85 \approx \boxed{39^\circ C}

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Mardokay Mosazghi
Jul 10, 2014

Heat lost by Aluminum = Heat gained by water. 'T' = final temp.

Al: 100 g x 0.897J/g/°C x (150 - T) ΔT. = 13,455 - 89.7T

H2O: 100 g x 4.18J/g/°C x (T - 15) ΔT. = 418T - 6,270.

13,455 - 89.7T = 418T - 6,270.

13,455 + 6,270 = 418T + 89.7T.

19,725 = 507.7T

T = 19,725 / 507.7 = 38.85°C final temp. (39°C)

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