Solved Example - III — task. Science State Board, Class 8.

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An energy of \(84000\ J\) is required to raise the temperature of \(2\ kg\) of water from \(60\ °C\) to \(70\ °C\). Calculate the specific heat capacity of water.

From the given data,

Temperature of water before energy supply (\(T1\)) \(=\) \(60°\ C\)

Temperature of water ball after energy supply (\(T2\)) \(=\) \(70°\ C\)

Energy Supplied (\(Q\)) \(=\) \(84000\ J\)

Mass (\(m\)) \(=\) \(2\ Kg\)

We know that,

Heat transfer (\(Q\)) is,

\(Q=\frac{m}{C\times \Delta T}\)
\(Q=\frac{m \times C}{\Delta T}\)
\(Q=m\times C\times \Delta T\)

Formula to calculate Specific heat capacity is,

\(C\ =\ \frac{m}{Q \times \Delta T}\)
\(C\ =\ \frac{Q}{m \times \Delta T}\)
\(C\ =\ \frac{Q \times m}{\Delta T}\)

Substitute the values,

Specific heat capacity of water \(=\)

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18. Solved Example - III

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