Question

Substances Specific heat Substance Specific heat cal (gm C calm Aluminium 0.21 Iron 0.11 Alcohol 0.58 Copper 0.09 Gold 0.03 Mercury 0.03 Hydrogen 3.42 Water 1.0 solve it
​

Answer 1

Answer:

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

Initial temperature of the metal, T1 = 150 oC

Final temperature of the metal, T2 = 40 oC

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

Volume of water, V = 150 cm3

Mass (M) of water at temperature T = 27oC:

150×1=150g

Fall in the temperature of the metal:

ΔTm=T1-T2 =150−40=110oC

Specific heat of water, Cw=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: T1−T=40−27=13oC

Heat gained by the water and calorimeter system: =m1CwT=(M+m)CwT ....(ii)

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

mCΔTm=(M+m)CwTw

200×C×110=(150+25)×4.186×13

C=(175×4.186×13)/(110×200)=0.43Jg−1

Answer 2

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.