Question

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 heatis required? What percentage of the heat is used to raise the temperature of (b) the panand (c) the water?

Answer 1

Answer:

heat used to rise temperature pan =  30.1%

heat used to rise temperature water =  69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature =  80.0ºC

to find out

heat used to rise temperature of  pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature    ...........1

specific heat of water is 4186 J/kgºC

so

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature    ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

so

total heat = 62.8 + 27 = 89.8 kJ

so

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

Answer 2

Heat raises the temperature of the pan and the water.

From the question, we know that;

mass of water = 0.250 liter = 0.250 kg

Aluminum pan mass = 0.500 kg

Initial temperature = 20.0ºC

Final temperature =  80.0ºC

The heat gained by water = 0.250 × 4186 × (80-20) kJ

=  62.8 kJ

Heat gained by the aluminum pan = 0.500 × 900 × (80-20) kJ

=  27 kJ

Total heat gained by water and pan = 62.8 kJ + 27 kJ =  89.8 kJ

Percentage of heat that raises the temperature of water = 62.8 / 89.8 ×100% = 69.9%

Percentage of heat that raises the temperature of the pan = 27/89.8 ×100% = 30.1%

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