Sensible Heat

Practice

practice problem 1

An 850 W consumer coffee maker can make 10 cups (1.75 liters) of 80 ℃ coffee from 20 ℃ tapwater in 10 minutes. What percentage of the electrical energy consumed actually makes it to the coffee?

solution

This is a problem about comparing the energy needed for some task (heating a pot of water) …

Qout = mcΔT = (1.75 kg)(4200 J/kg K)(80 − 20 ℃) = 441,000 J

to the energy consumed by the device responsible for getting the job done (electrical energy, but that doesn't really matter) …

Win = Pt = (850 W)(10 min)(60 s/min) = 510,000 J

The ratio of these two quantities is the efficiency of the device …

η = Qout/Win = (441,000 J)/(510,000 J) = 86.5%

practice problem 2

A 100 g cube of aluminum is removed from a bath of boiling water and dropped in a bath of room temperature water (2.00 liters at 20.0 ℃). What is the final temperature of the water and zinc assuming heat loss to the surroundings is negligible?

solution

This is a classic problem in conservation of energy. The heat lost by the hot object (in this case, the aluminum cube) equals the heat gained by the cold object (in this case, the water bath). Note how the order of subtraction affects the calculation of ΔT.

+Qhot  =  −Qcold
+(mcΔT)aluminum  =  −(mcΔT)water
(0.100 kg)(897 J/kg℃)(Tfinal − 100 ℃)  =  (2.00 kg)(4182 J/kg℃)(20.0 ℃ − Tfinal)
Tfinal  =  20.8 ℃

These results are quite sensible (no pun intended). The temperature of the water, which is more massive and has a higher specific heat, has barely changed at all.

practice problem 3

Write something different.

solution

Answer it.

practice problem 4

Write something completely different.

solution

Answer it.

The Physics Hypertextbook

  • No condition is permanent.