Question

5. 3.4 kg of water is heated from 78 C to boiling and it is all turned to steam at

100C. How much heat did it take?

Answer 1

7.7 MJ

Step-by-step explanation:

Let water specific heat be c = 0.004186 J/kgC and specific latent heat of vaporization be L = 2264705 J/kg

There would be 2 kinds of heat to achieve this:

- Heat to change water temperature from 78C to boiling point:

`H_1 = mc\Delta t = 3.4 * 0.004186 * (100 - 78) = 0.313 J`

- Heat to turn liquid water into steam:

`H_2 = mL = 3.4 * 2264705 \approx 7.7 * 10^6J` or 7.7 MJ

So the total heat it would take is
`H_1 + H_2 = 7.7 * 10^6 + 0.313 \approx 7.7 MJ`