How much heat energy in joules J is necessary to raise the temperature of $5$ kg of water from $20^{\circ} C$ to $100^{\circ} C$ ?

1674.4 k J

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167440 k J

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16720 k J

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16744 k J

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Solution

The correct option is C $16720 k J$
Given,
$m = 5 k g = 5, 000 g$
$Δ T = 1000 C - 200 C = 800 C$
The magnitude of heat energy in joule is $H$
$H = 4.186 \times 5000 \times 800 = 1672400 J$
$H = 1672 \times 10^{3} = 16720 K J$

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