A 100g aluminium calorimeter contains a mixture of 40 g of ice and 20 g of water

Question

A 100g aluminium calorimeter contains a mixture of 40 g of ice and 20 g of water at equilibriu. A copper cylinder of mass 300g is heated to 350 degrees celsius and then dropped into the calorimeter. What is the final temperature of the calorimeter and its contents if no hreat is lost to the surroundings? Express your answerusing three significant figures. Useful information : specific heat capacity of water= 1.00 cal/g C degree. Specific heat of aluminum= 0.22 cal/ g degree celsuis. Specific hear of copper= 0.093 cal/ C*g . Heat of fusion of water= 79.9 cal/g. (Answer=22.7 degrees celius)

Explanation / Answer

Aluminium calorimeter contains a mixture of 40g ice and 20g water at equilibrium. So it can be assumed that the temperature is 0 C

Here copper cylinder is at a higher temperature so it will loose heat and aluminium,water and ice will gain heat.

Heat lost by copper cylinder = mass x specific heat x temperature difference

= 300 x 0.093 x (350 - T) where T is the final equilibrium temperature

=9765 - 27.9 T

Heat gained = heat gained by aluminium + heat gained by water + heat gained by ice

= mass x specific heat of aluminium x temperature difference + mass x specific heat of waterx temperature difference + mass of ice x heat of fusion (for converting ice to water at 0 C) + mass of convrted water from ice x specific heat of water x temperature difference

= 100 x 0.22 x T + 20 x 1 x T + 40 x 79.9 + 40 x 1 x T

= 22 T +20 T + 3196 + 40 T

= 82 T + 3196

As heat lost = heat gain ( since no heat loss to the surroundings)

so,  9765 - 27.9 T = 82 T + 3196

so, T = 59.8 C

the answer for this question i.e. the final temperature of the calorimeter and its content is 59.8 C

the answer will be 22.7 degree celcius if water is 200 g and not 20g as mentioned in this question. (for solving the question with water 200g, please replace 20 with 200.)

all the best in the course work.

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