100 g of water is heated from 30°C to 50°C Ignoring the slight expansion of the water, the change in
its intemal energy is (specific heat of water is 4184 J/kg/K):-
[AIEEE - 2011)
(1) 84 kJ
(2)2.1 kJ
(3) 4.2 kJ
(4) 8.4 kJ
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On ignoring the slight expansion of the water, the change in its internal energy is 8.4 kJ.
Given-
- Weight of water = 100 gram
- Initial temperature = 30 °C
- Final temperature = 50 °C
- Specific heat of water = 4184 J/kg/K
In this case work done is zero.
W = PΔV and ΔV = 0 so W = 0
We know that change in internal energy is
ΔU = mCΔT
where m is the mass, C is the specific heat and T is the temperature.
By substituting the values we get
ΔU = 100 × 10⁻³ × 4184 × (50 -30)
ΔU ≈ 8.4 kJ
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