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Thermodynamics

Question : 1
Total: 10
Solution:
Here, mass of water heated, m=3000 g min1
Volume of water heated =3.0 lit min 1
Rise in temperature ΔT=(7727)°C=50°C
Specific heat of water c=4.2Jg1K1
Amount of heat used ΔQ=mcΔT
Now,
ΔQ
Δt
=
m
Δt
c
Δ
T

ΔQ
Dt
=
3000
1min
×4.2
×50
;

ΔQ
Δt
=63000Jmin1

Heat of consumption (ΔQ)=ΔmL
ΔQ
Δt
=
Δm
Δt
.L

630000 J min1
= (
Δm
Δt
)
×4
×104Jg1

Rate of consumption of fuel
Δm
Δt
=
63×104
4×104

=15.75 16g min1
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