Question

If you have ice cubes that are 10g each, calculate the minimum number of ice cubes (27°F) you would need to cool a 0.5 L cup of hot coffee (150°F) to a drinkable temperature (120°F). You may assume that coffee has the same density and thermal characteristics of water, and that there is no heat lost to the environment.

Answer 1

In the process of cooling a hot cup of coffe we use the following concept:

The amount of heat lost by coffee =amount of heat gained by coffee

(Assuming no heat is lost to the environment)

M-mass of coffee

S_h-specific of water

ΔT_coffee- change in temperature of coffee

n- no. of ice cubes

m- mass of one ice cube

L- latenet heat of fusion

ΔT_ice- changein temperature of ice

Now replacing each parameter with the numerical values

All temperatures are changed from fahrenheit to degree celsius

1.)Coffee

M=500 grams

S_h=1

ΔT_ice=65.55 - 48.88 = 16.67

2.) Ice

m = 10 grams

L = 80 calories/gram

ΔT_ice- 48.88-0

500X1X16.67=n(10X80 + 10X1X48.88)

n=6.46 ̴ 6

So, 6 ice cubes are required to cool a 0.5L cup of hot coffee (150 F) to a drinkable temperature (120 F).