An ice cube of mass of 10 g is taken from the freezer at-10ºC and placed into 20

Question

An ice cube of mass of 10 g is taken from the freezer at-10ºC and placed into 200 ml of water in a 100-g glasscontainer at 20ºC. What will be the final temperature of themixture after the ice has melted? The specific heat of ice is 2100J/(kg ºC), of water is 4186 J/(kg ºC) and of glass is 840J/(kg ºC). The latent heat of fusion of ice is 3.33 ×105 J/kg.
( show every stepplease) An ice cube of mass of 10 g is taken from the freezer at-10ºC and placed into 200 ml of water in a 100-g glasscontainer at 20ºC. What will be the final temperature of themixture after the ice has melted? The specific heat of ice is 2100J/(kg ºC), of water is 4186 J/(kg ºC) and of glass is 840J/(kg ºC). The latent heat of fusion of ice is 3.33 ×105 J/kg.
( show every stepplease)

Explanation / Answer

An ice cube of mass of 10 g is taken from the freezer at-10ºC and placed into 200 ml of water in a 100-g glasscontainer at 20ºC. What will be the final temperature of themixture after the ice has melted? The specific heat of ice is 2100J/(kg ºC), of water is 4186 J/(kg ºC) and of glass is 840J/(kg ºC). The latent heat of fusion of ice is 3.33 ×105 J/kg.
heat gained by ice during -10 to 0 degree=
10/1000 *2100*10=210J
heat gained by ice duringmelting=
10/1000 *333000=3330J
heat gained by convertedwater in reaching the resultant temp t (say)=10/1000 *4186*t J
total heat gained=210+3330+41.86t------(1)
heatlost by water while converting to water at t degree=
0.2*4186*(20-t)

heat lost by glass while converting to t degree=
0.1*840*(20-t)
total heatlost=0.2*4186*(20-t)+0.1*840*(20-t)----(2)
when there is no loss to the surroundings
heat lost=heatgained
0.2*4186*(20-t)+0.1*840*(20-t)=210+3330+41.86t
solving we will get the answer. t

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