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Amit Saxena Grade: upto college level
`        Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J/kg-K and its densities at 0°C and 4°C are 999.9 kg/m3 and 1000 kg/m3 respectively. Atmospheric pressure = 105 Pa. `
3 years ago

Navjyot Kalra
654 Points
```										Sol. Given M = 2 kg  2t = 4°c  Sw = 4200 J/Kg–k
f0 = 999.9 kg/m^3  f4 = 1000 kg/m^3  P = 10^5 Pa.
Net internal energy = dv
dQ = DU + dw ⇒ ms∆Q∅ = dU + P(v^0 – v^4)
⇒ 2 × 4200 × 4 = dU + 10^5(m – m)
⇒ 33600 = dU + 10^5 (m/V base 0 – m/v base 4) = dU + 10^5(0.0020002 – 0.002) = dU + 10^5 0.0000002
⇒ 33600 = du + 0.02 ⇒ du = (33600 – 0.02) J

```
3 years ago
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