Question

Density of water at 4\(^\circ\)C is 1000 kg/m\(^3\) and at 20\(^\circ\)C it is 998 kg/m\(^3\). If 4kg of water is heated from 4\(^\circ\)C to 20\(^\circ\)C, the change in internal energy of water is : (Given : specific heat capacity of water = 4200 J/kg K, Atmospheric pressure P = \(10^5\) Pa).

• A. 268799.2 J
• B. 268800.8 J
• C. 268800.0 J
• D. 267765.2 J

Hint:

For processes involving liquids and solids, the work done due to volume change is often very small compared to the heat supplied, but it's not always negligible, especially when high precision is required as in this question's options. Always calculate both Q and W unless you are sure W can be ignored.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are heating a mass of water, which causes its temperature, density, and volume to change. We need to find the change in its internal energy (\(\Delta U\)). This requires applying the First Law of Thermodynamics.
Step 2: Key Formula or Approach:
The First Law of Thermodynamics states:
\[ \Delta U = Q - W \] where \(\Delta U\) is the change in internal energy, \(Q\) is the heat supplied to the system, and \(W\) is the work done by the system.
- Heat supplied: \(Q = m s \Delta T\), where \(m\) is mass, \(s\) is specific heat capacity, and \(\Delta T\) is the change in temperature.
- Work done by the system (due to volume change against constant pressure): \(W = P \Delta V\), where \(P\) is the external pressure and \(\Delta V\) is the change in volume.
- Volume is related to mass and density by \(V = m/\rho\).
Step 3: Detailed Explanation:
Given values:
Mass, \(m = 4\) kg
Initial temperature, \(T_i = 4^\circ\)C; Final temperature, \(T_f = 20^\circ\)C
Change in temperature, \(\Delta T = T_f - T_i = 20 - 4 = 16^\circ\)C (or 16 K)
Specific heat, \(s = 4200\) J/kg K
Initial density, \(\rho_i = 1000\) kg/m\(^3\); Final density, \(\rho_f = 998\) kg/m\(^3\)
Pressure, \(P = 10^5\) Pa
Calculate Heat Supplied (Q):
\[ Q = m s \Delta T = 4 \times 4200 \times 16 \] \[ Q = 268800 \text{ J} \] Calculate Work Done (W):
First, find the change in volume \(\Delta V\).
Initial volume, \(V_i = m/\rho_i = 4 / 1000 = 0.004\) m\(^3\).
Final volume, \(V_f = m/\rho_f = 4 / 998\) m\(^3\).
\[ \Delta V = V_f - V_i = \frac{4}{998} - \frac{4}{1000} = 4 \left( \frac{1}{998} - \frac{1}{1000} \right) \] \[ \Delta V = 4 \left( \frac{1000 - 998}{998 \times 1000} \right) = 4 \left( \frac{2}{998000} \right) = \frac{8}{998000} \text{ m}^3 \] Now, calculate the work done:
\[ W = P \Delta V = 10^5 \times \frac{8}{998000} = \frac{800000}{998000} = \frac{800}{998} \] \[ W \approx 0.8016 \text{ J} \] Calculate Change in Internal Energy (\(\Delta U\)):
\[ \Delta U = Q - W = 268800 - 0.8016 \] \[ \Delta U = 268799.1984 \text{ J} \] Rounding to one decimal place, we get 268799.2 J.
Step 4: Final Answer:
The change in internal energy of water is 268799.2 J.