# Law of Conservation of Energy of a water heater

1. Jul 26, 2006

### quicknote

Imagine that your water heater has broken, but you want to take a bath. You fill your bathtub with 25 kg of room-temperature water (about 25 C). You figure that you can boil water on the stove and pour it into the bath to raise the temperature.
How much boiling water would you need in order to raise the bath to body temperature (about 37C)? Assume that no heat is transferred to the surrounding environment.

I know this is a simple question, but I'm totally stuck.

I calculated the amount of heat needed to raise the temperature to 37C:

q=Cm \Delta{T}
=1*25000*12
=300 000 cal.

Since energy is conserved, I used the the same equation to calculate the mass of the boiling water. The mass just equals 25kg, which is totally wrong.
Can anyone help me out and point me in the right direction?
Thank you.

2. Jul 26, 2006

### Staff: Mentor

Please show your work. Exactly what numbers did you plug in, and where? Then someone can probably tell you exactly what you did wrong.

3. Jul 26, 2006

### PPonte

We must assume that the boiling water is at 100 ºC.

When you pour the boiling water at the bathtub, you don't just transfer energy you add water too.

• We want the final temperature of water, $$T$$ to be: 37 ºC
• The initial temperature of the water in the bathtube, $$T_0$$ is 25 ºC.
• The initial temperature of the boiling water from the stove, $$T_s$$ is 100 ºC.
• We know also the mass of water in the bathtube, $$m_0$$ which is 25 kg.

The heat gained by the water in the bathtube is equal to the heat lost by the boiling water from the stove that was poured.

$$Q_{gain} = Q_{lost}$$

$$m_0 \times c_{water} \times (T - T_0) = m \times c_{water} \times (T_S - T)$$

$$m = \frac{m_0 \times (T - T_0)}{(T_S - T)}$$

4. Jul 26, 2006

### quicknote

Thanks PPonte!

Actually your first sentence helped me. I was using the wrong temperature

5. Jul 26, 2006

### PPonte

Anytime! I'm glad I could help.