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

1 Answer

3 votes

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*4.184*80 \\ 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

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