physics(specific latent heat)

2011-07-03 2:53 am

The problem is:
0.5 kg of ice at 0℃ is mixed with 1 kg of water at 20℃. Assume that there is no heat transfer between the mixture and the surroundings.Find the final temperature of the mixture.

Here is my working:
Let T be the final tmperature of the mixture.

Ice at 0℃ will change to water at the same temperature first and latent heat is absorbed. Then water at 0℃ will rise to a temperature of T.

Energy gained by the ice = Energy loss by the water
0.5*3.34*10^5+4200*0.5*(T-0)=4200*1*(20-T)
167000+2100T=84000-4200T
6300T=-83000
T=-13.174...
The final temperature is-13.2℃.

The correct answer is 0℃.

I know my answer is too unrealistic and I can't figure out how 0℃ can be calculated as the answer. So please tell me how to do the question or the mistakes made. Better with all the steps if you can show me. Thank you very much!

回答 (1)

天同

2011-07-03 3:43 am

✔ 最佳答案

Heat needed to melt all 0.5 kg of ice= 0.5 x (3.34x10^5) J = 167 000 J
Heat given out by the 1 kg water water when it is cooled to feezing point
= 1 x 4200 x 20 J = 84 000 J

By comparing the two results, you would realize that the geat released by the 1 kg water (84 kJ) is not sufficient to provide the required heat (167 kJ) to melt all 0.5 kg ice.

Hence, only part of the ice could be melted. The final mixture is therefore ice and water in wquilibrium, and the temperature is clearly at 0 degree Celsius.

The -ve temperature you obtained in your result just indicates that the heat given out by the water, when cooled from 20'C to 0'C, is not sufficient to melt all the ice. In fact, the equation to describe this process is,

(3.34x10^5)m = 1 x 4200 x 20
where m is the mass of ice melted, and surely is less than 0.5 kg.

👍 0 👎 0