Question

As an eager scientist on a hot summer day you wish to determine how much ice to buy to add to your cooler which is filled with 32 cans of soda which are warm at 82.9°F. Each can has a mass of 402 g and ideally you want the temperature of the drinks to be 40.1°F. If there is no heat lost by the cooler and ignoring any heat lost to the soda containers, how much ice needs to be added to your cooler? (Assume the temperature of the ice is 32.0°F.)

HINT: This problem is an example of calorimetry. The cooler acts like a calorimeter since there is not heat loss through it. Therefore, the amount of heat transferred to melt the ice and raise the temperature of the melted ice will be equal to the heat to cool the soda cans. The latent heat of fusion for ice is 80.0 cal/g, the specific heat of water is 1.00 cal/g·°C and the specific heat of the soda is 0.900 cal/g·°C and there are 454 grams per pound.

Answer #1

Solution :-

Ice will absorb heat and soda can will loose heat

32 soda can *402 g / 1 can = 12864 g

T of ice = 32 F = 0 C

T of soda can = 82.9 F = 28.28 C

T of the mixture = 40.1 F = 4.5 C

Lets make the following set up

-q soda can = q ice

- m*c*delta T = (m* delta H fus ) + (m*c*delta T)

-12864 g * 0.900 cal per g C* (4.5 C – 28.28 C) = (m*80.0 cal per g )+(m*1 cal per g * 4.5 C)

275315.328 cal = 80.0 m + 4.5 m

275315.328 cal = 84.5 m

275315.328 / 84 = m

3258 g = m

So the mass of ice needed = 3258 g

We can round it to 3260 g Ice

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