Two iron balls of different mass are heated to 100°C and dropped in water. If the same amount of heat is lost by the two balls to water, what can be said about the final temperatures of the two balls? (Heat lost = mCpΔT, where m = mass of the object, Cp = specific heat capacity of the material, and ΔT = change in temperature).

Question

Malaki Campbell

Chemistry

5 September, 13:24

Two iron balls of different mass are heated to 100°C and dropped in water. If the same amount of heat is lost by the two balls to water, what can be said about the final temperatures of the two balls? (Heat lost = mCpΔT, where m = mass of the object, Cp = specific heat capacity of the material, and ΔT = change in temperature).

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Donald · Answer 1

Since q (Heat lost) = mCpΔT, and the two balls lost the same amount of heat, we can show that m1CpΔT1 = m2CpΔT2. Since the balls are both made from iron and are at the same original temperature, Cp is the same in both cases. However, since m1 and m2 are different and Cp is constant, in order for the above expression to hold true, ΔT1 must also be different from ΔT2. Therefore each ball's temperature changes by a different amount, so their final temperatures are different.