Question

420 joule energy supplied to 10 gram of water will raise its temperature by​

Answer 1

The specific heat of water is 4.2 J/gm-K.

Given: mass of water = 10 grams

So, heat capacity = specific heat capacity * mass = 4.2 * 10 J/K = 42 J/K.

Heat supplied = Heat capacity * Increase in temperature, so:

420 J = 42 J/K * ΔT

ΔT = 10 K, or 10 °C