Question

I have a cup of hot coffee at 140 oC but I want to cool it to 110 oC. My cup holds about 0.3 kg of coffee. Fortunately, I have a bunch of aluminum cubes in the freezer that I can drop into my hot coffee to cool it down.

If each aluminum cube has a mass of 1 g (not 1 kg!) and my freezer keeps its contents at a temperature of –10 oC, how many cubes do I have to drop into my coffee? The specific heat of water is around 4000 joules/kg/oC and aluminum is about 900 joules/kg/oC. (Pick the answer closest to the true value and ignore any thermal losses to surroundings.)

A. 200
B. 330
C. 400
D. 110
E. 88

Answer 1

The correct answer is option B.

Step-by-step explanation:

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`C_1` = specific heat of metal =
`900 J/kg^oC`

`C_2` = specific heat of coffee=
`4000 J/kg^oC`

`m_1` = mass of metal = x

`m_2` = mass of coffee = 0.3 kg

`T_f` = final temperature of aluminum metal=
`110^oC`

`T_1` = initial temperature of aluminum metal =
`-10^oC`

`T_2` = initial temperature of coffee=
`140^oC`

Now put all the given values in the above formula, we get

`x* 900 J/kg^oC* (110-(-10))^oC=-(0.3 kg* 4000 J/kg^oC* (110-140)^oC`

`x=0.333 kg`

Mass of aluminum cubes = 0.3333 kg = 333.3 g

If mass of 1 cube is 1 gram, then numbers of cubes in 333.3 grams will be:

`=(333.3 g)/(1 g)=333.3\approx 330`

330 cubes of aluminum cubes will be required.