Some crushed ice, at 0 °C is dried quickly with a blotting paper and transferred to a known mass of water in an insulated container.
The following results were obtained.
Initial mass of water in beaker at the start = 100g
Mass of ice transferred to the beaker = 10g
Initial temperature of water = 23 °C
Final temperature of water = 16 °C
(a) Given that water has a specific heat capacity of 4200 J/(kg °C), calculate
(i) the heat energy released by the water in cooling,
(ii) the latent heat to melt the ice,
(iii) a value for the specific latent heat of fusion for ice.
(b) Would the value obtained for (a)(iii) likely to be higher or lower than the actual value? Explain your answer.
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Answer:
(a)
(i) Heat energy released by water = mcΔθ = 0.1 * 4200 * (23 - 16) = 2940 J
(ii) Heat energy released by water = latent heat to melt the ice + heat energy gained by water (from melted ice)
2940 = latent heat to melt the ice + 0.01 * 4200 * (16 - 0)
2940 = latent heat to melt the ice + 672
latent heat to melt the ice = 2940 - 672 = 2268 J
(iii) Let the specific latent heat of fusion for ice be L
latent heat to melt the ice = mL = 0.01 L
2268 = 0.01 L
L = 226800 J / kg
(b) It would be higher.
This is because heat is also released by the beaker,
i.e.Heat energy released by water + Heat released by beaker = latent heat to melt the ice + heat energy gained by water (from melted ice)
Thus, latent heat to melt the ice = Heat energy released by water + Heat released by beaker - heat energy gained by water (from melted ice)
Note: You cannot say heat is gained (or lost) to the surroundings, because it is already stated that the water is in an insulated container.
