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Calculating the required heat: heating water in an aluminum pan

A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC . (a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water?

Strategy

The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and the pan is increased by the same amount. We use the equation for the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for water and aluminum are given in [link] .

Solution

Because water is in thermal contact with the aluminum, the pan and the water are at the same temperature.

  1. Calculate the temperature difference:
    Δ T = T f T i = 60.0ºC.
  2. Calculate the mass of water. Because the density of water is 1000 kg/m 3 , one liter of water has a mass of 1 kg, and the mass of 0.250 liters of water is m w = 0 . 250 kg size 12{m rSub { size 8{w} } =0 "." "25"`"kg"} {} .
  3. Calculate the heat transferred to the water. Use the specific heat of water in [link] :
    Q w = m w c w Δ T = 0 . 250 kg 4186 J/kg ºC 60.0 ºC = 62 .8 kJ.
  4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in [link] :
    Q Al = m Al c Al Δ T = 0.500 kg 900 J/kgºC 60.0ºC = 27.0 × 10 4 J = 27.0 kJ.
  5. Compare the percentage of heat going into the pan versus that going into the water. First, find the total transferred heat:
    Q Total = Q W + Q Al = 62 . 8 kJ + 27 . 0 kJ = 89 . 8 kJ. size 12{Q rSub { size 8{"Total"} } =Q rSub { size 8{W} } +Q rSub { size 8{"Al"} } ="62" "." 8`"kJ "+" 89" "." 5`"kJ = 152" "." 3`"kJ"} {}

Thus, the amount of heat going into heating the pan is

27 . 0 kJ 89 . 8 kJ × 100% = 30.1%, size 12{ { {"62" "." 8`"kJ"} over {"152" "." 3`"kJ"} } times "100"%="41"%} {}

and the amount going into heating the water is

62 . 8 kJ 89 . 8 kJ × 100% = 69.9% . size 12{ { {"62" "." 8`"kJ"} over {"89" "." 8`"kJ"} } times "100"%="69.9"% "." } {}

Discussion

In this example, the heat transferred to the container is a significant fraction of the total transferred heat. Although the mass of the pan is twice that of the water, the specific heat of water is over four times greater than that of aluminum. Therefore, it takes a bit more than twice the heat to achieve the given temperature change for the water as compared to the aluminum pan.

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The figure shows a truck coming from the left and moving on a road which is sloping downhill to the right. Smoke is coming from the area of the wheels of the truck.
The smoking brakes on this truck are a visible evidence of the mechanical equivalent of heat.

Calculating the temperature increase from the work done on a substance: truck brakes overheat on downhill runs

Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the brake material. This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck. The problem is that the mass of the truck is large compared with that of the brake material absorbing the energy, and the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment.

Calculate the temperature increase of 100 kg of brake material with an average specific heat of 800 J/kg ºC if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed.

Strategy

If the brakes are not applied, gravitational potential energy is converted into kinetic energy. When brakes are applied, gravitational potential energy is converted into internal energy of the brake material. We first calculate the gravitational potential energy ( Mgh ) size 12{ \( ital "Mgh" \) } {} that the entire truck loses in its descent and then find the temperature increase produced in the brake material alone.

Practice Key Terms 1

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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