# Heat transffer and specific heat: Mixture method

A 150 gram aluminum container contains 100 grams of water at 30 C, 50 grams of copper is add at 100 C and 40 grams of ice at 0 C. How do I calculate the final temperature of this mixture?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

In such problems where change in state occurs the main problem is to decide the configuration of the final mixture. As in this problem we have to find that whether the final mixture is having ice and water or only water i.e. whether the whole ice will melt or not. This can be decided by calculating the amount of heat required to change the state and the heat available for the same.

At first we have decide the configuration of final mixture.

Heat required to melt the whole ice = mL= 40 x 333 =13320 J.(letant heat of melting of ice is 333 J/g)

If the whole ice is not melted the temperature of mixture will be 0 C

Heat given by the container, water and copper to cool down to 0 C

= 150 x 0.90 x (30-0) + 100 x 4.18 x (30-0) + 50 x 0.39 (100 - 0)

= 4050 + 12540 + 1950

= 18540 J

This is grater then the heat required to melt the whole ice 13320 J hence the whole ice will be melted and still some heat is available to increase the temperature.

So the whole ice will get melted and the temperature of the mixture will be greater then 0 C.

Let the final temperature of the mixture be t C

Heat required for melting ice and increasing the temp of water produced by its melting

= 13320 + 40 x 4.18 x (t - 0) J

Heat given out by container, water and copper to cool down to t C

= 150 x 0.90 x (30-t) + 100 x 4.18 x (30-t) + 50 x 0.39 (100 - t)

= 18540 - t(135 + 418 + 19.5)

According to law of mixture

13320 + 40 x 4.18 x (t - 0) J = 18540 - t(135 + 418 + 19.5)

gives t = 9.12 C

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