Question

Suppose you raise the temperature of \(1 \mathrm{~kg}\) of water from \(-10^{\circ} \mathrm{C}\) to \(+10^{\circ} \mathrm{C},\) then from \(+20^{\circ} \mathrm{C}\) to \(+40{ }^{\circ} \mathrm{C},\) and finally from \(+90{ }^{\circ} \mathrm{C}\) to \(+110{ }^{\circ} \mathrm{C}\) Which of the three changes in temperature requires the addition of the least heat? a) first change b) second change c) third change d) All of the changes require the same amount of heat.

Short answer

Answer: d) All of the changes require the same amount of heat.

Step-by-step solution

Find the change in temperature for each case

For the first change, we have a temperature change from \(-10^{\circ} \mathrm{C}\) to \(+10^{\circ} \mathrm{C}\). The change in temperature is given by: \(ΔT_1 = \mathrm{Final \: Temperature} - \mathrm{Initial \: Temperature} = 10 - (-10) = 20^{\circ} \mathrm{C}\). For the second change, we have a temperature change from \(+20^{\circ} \mathrm{C}\) to \(+40^{\circ} \mathrm{C}\). The change in temperature is given by: \(ΔT_2 = 40 - 20 = 20^{\circ} \mathrm{C}\). For the third change, we have a temperature change from \(+90^{\circ} \mathrm{C}\) to \(+110^{\circ} \mathrm{C}\). The change in temperature is given by: \(ΔT_3 = 110 - 90 = 20^{\circ} \mathrm{C}\).

Compare the changes in temperature

Now, we have calculated the changes in temperature for all the three cases: \(ΔT_1 = 20^{\circ} \mathrm{C}\), \(ΔT_2 = 20^{\circ} \mathrm{C}\), \(ΔT_3 = 20^{\circ} \mathrm{C}\). Since the mass and specific heat capacity of water are constant for all cases, the amount of heat required for each case depends only on the change in temperature. As we see, all three changes in temperature are equal, and thus the amount of heat required for each case is the same. Therefore, the correct answer is: d) All of the changes require the same amount of heat.