The specific heat of milk is 3.93 103 J/(kg · K), and the specific heat of water

Question

The specific heat of milk is 3.93 103 J/(kg · K), and the specific heat of water is 4.19 103 J/(kg · K). Suppose you wish to make a large mug (0.500 L) of hot chocolate. Each liquid is initially at 5.20°C, and you need to raise their temperature to 75.0°C. The density of milk is about 1.03 103 kg/m3, and the density of water is 1.00 103 kg/m3. a) How much heat must be transferred in each case?

(b) If you use a small electric hot plate that puts out 485 W, how long would it take to heat each liquid?

Explanation / Answer

density of milk=1.03*10^3 kg/m^3

volume=0.5 L=0.5*10^(-3) m^3

mass of milk=density*volume=1.03*10^3*0.5*10^(-3)=0.515 kg.

specific heat of milk=3.93*10^3 J/(kg.K)

then heat transfer required=mass*specific heat*temperature difference

=0.515*3.93*10^3*(75-5.2)

=1.4127*10^5 J

density of water=1.00*10^3 kg/m^3

volume=0.5 L=0.5*10^(-3) m^3

mass of water=density*volume=1.00*10^3*0.5*10^(-3)=0.5 kg.

specific heat of water=4.19*10^3 J/(kg.K)

then heat transfer required=mass*specific heat*temperature difference

=0.5*4.19*10^3*(75-5.2)

=1.4623*10^5 J

part b:

power output=485 W

then time required to heat milk=energy required/power=1.4127*10^5/485=291.28 seconds

time required to heat water=energy required/power=1.4623*10^5/485=301.51 seconds

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