At 298k the coefficient of thermal expansion of water is α=2.07 *10-⁴ ...
Calculation of work, heat and change in internal energy of water
Given
- Coefficient of thermal expansion of water (α) = 2.07 * 10^-4 K^-1
- Density of water (ρ) = 0.9970 g/m³
- Initial temperature (T1) = 298 K
- Final temperature (T2) = 323 K
- Constant pressure (P) = 101 kPa
- Mass of water (m) = 200 g
- Specific heat capacity of water at constant pressure (CP) = 75.30 J/K.mol
Calculation of Work (W)
The work done by the system is given by the formula:
W = -PΔV
where,
- P = Constant pressure = 101 kPa
- ΔV = Change in volume
Change in volume (ΔV) can be calculated using the formula:
ΔV = V2 - V1
where,
- V1 = Initial volume = m/ρ = 200/0.9970 = 200.80 cm³
- V2 = Final volume
Final volume (V2) can be calculated using the formula:
V2 = V1 (1 + αΔT)
where,
- ΔT = Change in temperature = T2 - T1 = 323 - 298 = 25 K
Substituting the values in the above equation, we get:
V2 = 200.80 (1 + 2.07 * 10^-4 * 25) = 201.79 cm³
Now, substituting the values of P, ΔV and converting the units to joules, we get:
W = -101 * 10^3 * (201.79 * 10^-6 - 200.80 * 10^-6) = -0.203 J
Calculation of Heat (Q)
The heat absorbed by the system is given by the formula:
Q = mCPΔT
where,
- m = Mass of water = 200 g
- CP = Specific heat capacity of water at constant pressure = 75.30 J/K.mol
- ΔT = Change in temperature = T2 - T1 = 323 - 298 = 25 K
Substituting the values in the above equation, we get:
Q = 200/18 * 75.30 * 25 = 20925 J
Calculation of Change in Internal Energy (ΔU)
The change in internal energy (ΔU) is