Vikas TU
Last Activity: 5 Years ago
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