How much heat is required to raise the temperature of 85 grams of water from $280 K$ to $342 K$ ?

5270 J

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355 J

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259 J

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151 J

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22029 J

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Solution

The correct option is E $22029 J$
$q = m S Δ T$
$q =$ heat required
$m = 85$ grams = mass of water
$S = 4.18$ J/gC = specific heat of water
$Δ T = 342 - 280 = 62$ K
$q = 85 \times 4.18 \times 62 = 22029 J$

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