A 80-g ice cube at 0°C is placed in 800 g of water at 24°C. What is the final temperature of the mixture?

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gsenviro | College Teacher | (Level 1) Educator Emeritus

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80 gm ice cube at 0 degree C and 800 gm water at 24 degree C.

We can start by understanding the basic processes that will take place. Heat flows from hot body to cold body. So, heat will flow from water to ice cube and will melt it. After that heat will flow from hot water to cold water, till an equilibrium temperature is reached.

Heat of fusion of ice = 333.55 KJ/kg

Amount of heat required to melt 80 gram of ice = 333.55 x 80/1000 KJ = 26.684 KJ

This is also the amount of heat lost by 800 gm of water, causing drop in its temperature.

Specific heat of water = 4.18 KJ/kg/degree C (or 4.18 J/g/C)

Thus, 26.684 KJ = 800/1000 kg x 4.18 KJ/kg/C x (24-T)

solving this, we get, T = 16.02 degree C

Now we have 80 grams of water at 0 degree C and 800 gm of water at 16.02 degree C. Using the fact that heat lost from hotter water will go to colder water,

80 x 4.18 x (Tf-0) = 800 x 4.18 x (16.02-Tf)

Solving this, we get: Tf = 12.93 degree C.

Hope this helps.