Question

2 kg of iron at 80°C is placed in a copper can of mass 0.5kg containing 1 kg of water at 20°C. After stirring, the temperature of the mixture is 30°C. Calculate the specific heat capacity of iron
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Answer 1

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Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, T

1

= 150

o

C

Final temperature of the metal, T

2

= 40

o

C

Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g

Volume of water, V = 150 cm

3

Mass (M) of water at temperature T = 27

o

C:

150×1=150g

Fall in the temperature of the metal:

ΔT

m

=T

1

-T

2

=150−40=110

o

C

Specific heat of water, C

w

=4.186J/g/K

Specific heat of the metal =C

Heat lost by the metal, =mCT .... (i)

Rise in the temperature of the water and calorimeter system: T

1

−T=40−27=13

o

C

Heat gained by the water and calorimeter system: =m

1

C

w

T=(M+m)C

w

T ....(ii)

Heat lost by the metal = Heat gained by the water and colorimeter system

mCΔT

m

=(M+m)C

w

T

w

200×C×110=(150+25)×4.186×13

C=(175×4.186×13)/(110×200)=0.43Jg

−1

k

−1

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

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Answer 2

Step-by-step explanation:

Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, T

1

= 150

o

C

Final temperature of the metal, T

2

= 40

o

C

Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g

Volume of water, V = 150 cm

3

Mass (M) of water at temperature T = 27

o

C:

150×1=150g

Fall in the temperature of the metal:

ΔT

m

=T

1

-T

2

=150−40=110

o

C

Specific heat of water, C

w

=4.186J/g/K

Specific heat of the metal =C

Heat lost by the metal, =mCT .... (i)

Rise in the temperature of the water and calorimeter system: T

1

−T=40−27=13

o

C

Heat gained by the water and calorimeter system: =m

1

C

w

T=(M+m)C

w

T ....(ii)

Heat lost by the metal = Heat gained by the water and colorimeter system

mCΔT

m

=(M+m)C

w

T

w

200×C×110=(150+25)×4.186×13

C=(175×4.186×13)/(110×200)=0.43Jg

−1

k

−1

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value