Question

Brewed coffee is often too hot to drink right away. You can cool it with an ice cube, but this dilutes it. Or you can buy a device that will cool your coffee without dilution - a 250 g aluminum cylinder that you take from your freezer and place in a mug of hot coffee. If the cylinder is cooled to -20∘C , a typical freezer temperature, and then dropped into a large cup of coffee (essentially water, with a mass of 500 g ) at 85∘C , what is the final temperature of the coffee?

Answer 1

Given mass of aluminum =250g

initial temperature of aluminum =$-20^{\circ}C$

mass of coffee = 500g

initial temperature of coffee =$85^{\circ}C$

specific heat of water = 4.18 J/g$^{\circ}C$

specific heat of aluminum =0.902J/g$^{\circ}C$

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when the cold cylinder is dropped into hot coffee,coffee losses heat and cools

This lost heat is gained by aluminum cylinder until temperature equalises

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so Heat lost by coffee =Heat gained by aluminum

$Q_{lost}=Q_{gained}$

$m_{coffee}c_{water}\Delta T_{coffee}=m_{Al}c_{Al}\Delta T_{Al}$

$500g*4.18J/g^{\circ}C*(85^{\circ}C-T_{f})=250g*0.902J/g^{\circ}C*(T_{f}--20^{\circ}C)$

$500*4.18*(85-T_{f})=250*0.902*(T_{f}--20)$

$500*4.18*(85-T_{f})/(250*0.902)=(T_{f}+20)$

$9.268*(85-T_{f})=(T_{f}+20)$

$9.268*85-9.268T_{f}=T_{f}+20$

$9.268*85-20=10.268T_{f}$

$767.78=10.268T_{f}$

ANSWER: $T_{f}=74.77^{\circ}C$

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