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note that the latent heat of phase change and the change and the specific heat capacity must be considered.

MASS OF ICE CUBES TEMPERATURE OF ICE CUBES a) 20g -6.0 deg c b) 10g -12 deg c c) 40g -3.0 deg c

now answer these questions. (i) what is the effectiveness of cooling by sample (a)? (ii) what is the effectiveness of cooling by sample (b)? (iii) what is the effectiveness of cooling by sample (c)?

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Let us have a mass 100 gm of water at 40 deg C, in contact with the ice cubes in each case.

specific heat capacity of ice = 2.11 J/deg C/gm Latent heat of fusion of ice = 334 J/gm specific heat of water = 4.18 J/gm/degC

Heat energy that these ice cubes will take to reach 0 deg C and then melt at 0 deg C and then heat to T deg C is :

a) 20 gms at -6 deg C 20 * 2.11 * 6 + 20 * 334 + 20 * 4.18 * T = 6, 933.2 + 83.6 T Joules = 100 * 4.18 (40-T) => T = 19.5 deg C

b) 10gm at -12 deg C 10 * 2.11 * 12 + 10 * 334 + 10 * 4.18 * T = 3,593.2 + 41.8 T Joules = 100 * 4.18 (40-T) => T = 28.5 deg C

c ) 40 gm at -3 deg C 40 * 2.11 * 3 + 40 * 334 + 40 * 4.18 * T = 13, 613.2 + 167.2 T Joules = 100 * 4.18 (40-T) => T = 5.3 deg C

Obviously the 40 gm ice cubes will be more effective. Although the heat required to heat the Ice to zero deg C is same, the latent heat needed is highest for the 40 gm sample.