Test Index

## Thermodynamics

Question : 1

Total: 10

**Solution:**

Here, mass of water heated, m = 3000 g m i n − 1

Volume of water heated= 3.0 lit min − 1

Rise in temperatureΔ T = ( 77 − 27 ) ° C = 50 ° C

Specific heat of waterc = 4.2 J g − 1 K − 1

Amount of heat usedΔ Q = m c Δ T

Now,

=

c Δ T

⇒

=

× 4.2 × 50 ;

= 63000 J min − 1

Heat of consumption( Δ Q ) = Δ m L

=

. L

⇒ 630000 J m i n − 1

= (

) × 4 × 10 4 J g − 1

Rate of consumption of fuel

=

= 15.75 ≈ 16 g min − 1

Volume of water heated

Rise in temperature

Specific heat of water

Amount of heat used

Now,

Heat of consumption

Rate of consumption of fuel

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