(i) the heat capacity of 4.0 kg of liquid, and
(ii) the specific heat capacity of liquid
Solution:
According to the statement, Power (P) of heater is 600 W
Mass (m) of liquid is 4.0 kg
Change in temperature of liquid becomes
(15 – 10)0C = 50 C (or 5 K)
And the time taken to raise the heater to this temperature is 100 s
We know that the expression for heat energy is => △Q = P × t
Putting vales and solving, △Q = 600 × 100
We get △Q = 60000 J
Also, the heat energy can be expressed by the equation => △Q = mc△T
Upon re-arranging, we have ==> c = △Q / m△T
Putting vales and solving, c = 60000 / (4 × 5)
c = 3000 J kg-1 K-1 = 3 × 103 J kg-1 K-1
And the Heat capacity is given by the relation
Heat capacity = c × m
Therefore, Heat capacity = 4 × 3000 J kg-1 K-1 = 1.2 × 104 J / K