(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