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A block of aluminum of mass 2 kg was supplied with 3600 J of heat. If the specific heat of aluminum is $900 {JKg}^{- 1} {°C}^{- 1}$ then the rise in temperature is

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3°C

2°C

4°C

5°C

answer is B.

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Detailed Solution

m = 2 kg,Q = 3600 J,S = 900 {JKg}^{- 1} ° C^{- 1}, Δ t = ?

$Q = ms Δ t$

$Δ t = \frac{Q}{ms} = \frac{3600}{2 \times 900} = 2 ° C$

$Δ t = 2 ° C$

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