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SPECIFIC HEAT PRACTICE PROBLEM AND ANSWER

 

1.     A pot is filled with 5 kg of water at 20oC. Calculate how much heat energy would be needed to raise the temperature to 70oC. [specific heat capacity of water = 4200J/kgoC ]

Answer :

Given : m = 5 kg

            Cwater = 4200J/kg0C

            ΔT = 700C – 200C = 500C

Find : q = ….?

Formula : q = m c ΔT

                    =  5 kg x 4200J/kgoC x 50oC

                    = 1050000 J

                    = 1.05 MJ

2.     How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 24°C to 55°C, if the specific heat of aluminum is 0.90 J/g°C

Answer :

Given : m = 10 g = 0.01 kg

            Caluminum = 0.90 J/kg0C

            ΔT = 55oC – 240C = 31oC

Find : q = ….?

Formula : q = m c ΔT

                    = 0.01 kg x 0.90 J/kgoC x 31oC

                    = 0.279 J

3.     Calculate the heat capacity of a piece of wood if 300.0 g of the wood absorbs 1.35×104 Joules of heat and its temperature changes from 24°C to 49°C.

Answer :

Given : m = 300 g = 0.3 kg

             q= 1.35 x 104 J

            ΔT = 49oC – 240C = 25oC

Find : c = ….?

Formula : c = q : ( m  ΔT )

                   = 1.35 x 104J : ( 0.3 kg x 25oC )

                   = 1.8 x 103 J/kgoC

 

4.     The temperature of 250 g of water changed from 24oC to 52oC. How much heat did this sample absorb? Cwater = 4.18 J/goC

Answer :

Given : m = 250 g

            Cwater = 4.18 J/g0C

            ΔT = 52oC – 240C = 28oC

Find : q = ….?

Formula : q = m c ΔT

                    = 250g x 4.18 J/goC x 28oC

                    = 29260 J

                    = 29.26 kJ

                   

5.     What is the specific heat of silicon if it takes 576 J to raise the temperature of 45.0g of Silicon by 18.0oC?

Answer :

Given : m = 45.0 g

             q= 576 J

            ΔT = 18oC

Find : c = ….?

Formula : c = q : ( m  ΔT )

                   = 576J : ( 45 g x 18oC)

                   = 0.71 J/goC

 

6.     What is the specific heat of lead if it takes 32 J to raise the temperature of a 25g block by 10oC?

Answer :

Given : m = 25 g

             q= 32 J

            ΔT = 10oC

Find : c = ….?

Formula : c = q : ( m  ΔT )

                   = 32 J : ( 25 g x 10oC )

                   = 0.128 J/goC

 

7.     250.0g of 22°C water is heated until its temperature is 37°C.  If the specific heat of water is 4.18 J/g°C, calculate the amount of heat energy needed to cause this rise in temperature.

Answer :

Given : m = 250 g

            Cwater = 4.18 J/g0C

            ΔT = 37oC – 220C = 15oC

Find : q = ….?

Formula : q = m c ΔT

                    = 250g x 4.18J/goCx15oC

                    = 15 675 J

 

8.     A 0.25 g piece of copper is heated and fashioned into a bracelet. The amount of energy transferred by heat to the copper is 1950 J. If the specific heat of copper is 390 J/goC, what Is the change of the copper's temperature?

Answer:

Given : m = 0.25 g

            Ccopper = 390 J/kg0C

            q = 1950J

Find : ΔT = ….?

Formula : ΔT  = q: (m c )

                        = 1950J : ( 0.25 g x 390J/goC )

                        = 20oC

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