Physics

How much heat energy is released when 500 g of water at 80° C cools down to 0° C and then completely freezes? [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) = 500 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 80° C to 0° C

= m x c x change in temperature

Substituting the values in the formula we get,

Q1 = 500 x 4.2 x (80 - 0)

= 500 x 4.2 x 80

= 168 x 103 J

(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 = 500 x 336

= 168 x 103 J

From relation,

Q = Q1 + Q2

= 168 x 103 J + 168 x 103 J

= 336 x 103 J

Total heat released = 336 x 103 J

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