Question

A liquid of specific heat capacity 3J/g K or 300J/kg K rises from a temperature of 15C to 65C in 1 min when an electric heater is used.if the heater generates 63 kilojoules per minute,calculate the mass of a liquid​

Answer 1

Answer:

Answer

Power of heater P=10 W

Let the specific heat of container be S

c

and that of oil be S

o

.

Let mass of container be m

c

.

For water-container system :

Given : m

w

=0.5 kg ΔT=3

o

K t=15×60=900 s S

w

=4200 Jkg

−1

K

−1

Heat given Pt=m

w

S

w

ΔT+m

c

S

c

ΔT

∴ 10(900)=0.5(4200)(3)+m

c

S

c

(3) ⟹m

c

S

c

=900 JK

−1

......(1)

For oil-container system :

Given : m

o

=2 kg ΔT=2

o

K t=20×60=1200 s

Heat given Pt=m

o

S

o

ΔT+m

c

S

c

ΔT

∴ 10(1200)=(2)S

o

(2)+(900)(2)

⟹S

o

=2.55×10

3

Jkg

−1

K

−1