In a calorimeter, 1 kg water is filled at room temperature(10°C). A water heater of negligible heat capacity having resistance for 400 ohm and carrying a current of 1 amp is dipped in water for 120 sec , due to which temperature of water and calorimeter is increased by 8°. Now in place of water an unknown liquid of mass 1kg is filled in the calorimeter at room temperature (10°C). A hot sphere of temperature 80° mass 2kg and specific heat 1/2 cal/gm°C is dipped in the liquid. If the final temperature of the mixture is found to be 50°C, then specific heat capacity of the unknown liquid is(in Cal/gm°C) : (J =4.2 J/cal)(Neglect heat loss to the surrounding).
Answers
Answer:
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