How many calories are needed to heat 100g of water from 278K to 288K ?

10 calories

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60 calories

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100 calories

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1000 calories

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Solution

The correct option is C 1000 calories
$q = m (c) Δ T$
$∴ q = (100 g) \times \frac{1 c a l}{(g - c)} \times (288 - 278) = 100 \times 10 c a l o r i e s = 1000 c a l o r i e s$

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100 g

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