When $60$ calories of heat is supplied to $15 g$ of water, the rise in temperature is

75^{\circ} C

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900^{\circ} C

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4^{\circ} C

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{0.25}^{\circ} C

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Solution

The correct option is C $4^{\circ} C$
The correct answer is option(C).
Given:
mass, $m = 15 g$ , heat supplied, $Q = 60 c a l$ , Specific heat of water, $C = 1 c a l g^{- 1} ° C^{- 1}$
We know,
$Q = m C Δ T ⟹ Δ T = \frac{Q}{m C} = \frac{60}{15 \times 1} = 4 ° C$
is the required rise in temperature.

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When 60 calories of heat is supplied to 15g of water, the rise in temperature is

The correct option is C 4∘CThe correct answer is option(C).Given:mass, m=15g, heat supplied, Q=60cal, Specific heat of water, C=1calg−1°C−1We know, Q=mCΔT⟹ΔT=QmC=6015×1=4°Cis the required rise in temperature.

When $60$ calories of heat is supplied to $15 g$ of water, the rise in temperature is