Question

What amount of energy is required to heat 48 grams of water from 40 °C to 120 °C?

Answer 1

Answer:

16,066.56Joules

Explanations:

The amount of heat energy required is expressed according to the formula;

$Q=mc\triangle t$

where:

• m is the mass of water = 48grams

,

• c is the specific heat capacity of water = 4.184 J/g°C

,

• △,t ,is the change in temperature, = 120 - 4,0 = 80°C

Substitute the given parameters into the formula

$\begin{gathered} Q=48\times4.184\times80 \\ Q=16066.56Joules \end{gathered}$

Hence the amount of energy required to heat 48 grams of water from 40 °C to 120 °C is 16,066.56Joules