Question

A cup containing 200g of water is sitting on your dining room table. After carefully measuring its temperature to be 20^oC, you leave the room. Returning ten minutes later, you measure its temperature again and find that it is now 25^oC. What can you conclude about the amount of heat added to the water? (Hint: This is a trick question.)

Short answer

Amount of heat added is 4186 J.

Step-by-step solution

Given information

Mass of water, m=200 g = 0.2 kg
Difference in temperature, ΔT=(25-20)^oC = 5^oC
Specific of water, C=4186 J /kgK

Explanation

Amount of heat added can be calculated as

Q=m c ΔT

Substitute values and calculate

$$\begin{gathered} Q=mc∆T \\ =(0.2k\mathrm{g})\times (4186\mathrm{J}/k\mathrm{g}°\mathrm{C})\times (5°\mathrm{C}) \\ Q=4186\mathrm{J} \end{gathered}$$

4186 Joule is added to the system.