Question

You took two 48 g ice cubes from your -13 C kitchen freezer and placed them in 200 g of water in a thermally insulated container. If the water is initally at 24 C, what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used?

Answer 1

total mass of ic e= 96 gm = 96 x 10^-3 kg

C ( of ice) = 2090 J/kg C, C of water ( 4182)

Heat lost by water = 200x 10^-3 ( 4182) ( 24) = 20073.6 J

Heat required by ice to reach 0 degree = 96 x 10^-3 ( 2090)(13) =2608.32

Which implies that remaning heat = 20073.6 J - 2608.32 J = 17465.28 J will be used to convert ice to water

latent heat of fision of ice= 334X 10³ (J/kg)

m 334 X 10³ (J/kg)= 17465.28

m = 52.29 gm of ice will convert to water at 0

final temp of thermal equilbrim = 0 ( some water + some ic e)

b) heat required by 48 g to reach 0 degree = 48 x 10^-3 ( 2090)(13) = 1304 . 16 J

Heat required to melt ice= 48 x 10^-3 334 X 10³ (J/kg)= 16032 J

Total heat = 17336.16 J

17336. 16 + 48 x 10^-3 ( 4182 ) ( T) =200x 10^-3 ( 4182) ( 24-T)

17336. 16 + 200.736 T = 20073.6 - 836.4 T

2737.44 = 1037.136 T

T (final Temp) = 2.639 C apprx