You would like to cool down a 360 ml can of soda from 30 degree C to 10 degree C

Question

You would like to cool down a 360 ml can of soda from 30 degree C to 10 degree C as quickly as possible in a salt water bath at -5 degree C. The can is a cylinder that has a diameter of 6.6 cm and is 12.0 cm high with very thin alumina side walls. Assume that the thermal properties of the soda and salt water are similar to those of water. Determine the total heat to be removed from the can. Assuming that the can is placed in the salt water ice bath and left alone, how long will it take to cool to desired temperature Your friend suggests that you swirl the can around in the water bath to improve the rate of cooling. Assume that the can is moving at rate equivalent to 0.1 m/s in the water and the soda is well mixed, how long will it take for the contents

Explanation / Answer

(i) q =mcpdT= 0.36*4.2(30-10)=30.24kJ

(ii) heat rejected by water= heat transferred to salt water

-hA(T-To)=mcpdT/dt

h=1.52(20)*1/3=10.13

A=1/3*phi*r2*h=136.77cm2

-10.13*136.77*10-4(30-(-5))=0.36*4200*(30-10)/dt

dt=6235.05sec

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