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How much heat energy is released when 5.0 g of water at 20° C changes into ice at 0° C? Take specific heat capacity of water = 4.2 J g-1 K-1, specific latent heat of fusion of ice = 336 J g-1.

Calorimetry

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Answer

Given,

Mass (m) = 5.0 g

Specific heat capacity of water (c) = 4.2 J g-1 K-1

Specific latent heat of fusion of ice (L) = 336 J g-1

(i) Heat energy released when water lowers it's temperature from 20° C to 0° C

= m x c x change in temperature

Substituting the values in the formula we get,

Q1=5×4.2×(200)Q1=5×4.2×20Q1=420JQ1 = 5 \times 4.2 \times (20 - 0) \\[0.5em] Q1 = 5 \times 4.2 \times 20 \\[0.5em] \Rightarrow Q_1 = 420 J \\[0.5em]

(ii) Heat energy released when water at 0° C changes into ice at 0° C = m x L

Substituting the values in the formula we get,

Q2=5×336JQ2=1680JQ2 = 5 × 336 J \\[0.5em] \Rightarrow Q2 = 1680 J \\[0.5em]

From relation,

Q=Q1+Q2Q=420+1680Q=2100Q = Q1 + Q2 \\[0.5em] Q = 420 + 1680 \\[0.5em] \Rightarrow Q = 2100 \\[0.5em]

Total heat released = 2100 J

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