Question

Water of mass 5 kg in a closed vessel is at a temperature of \(20 \, ^\circ\text{C}\). If the temperature of the water when heated for a time of 10 minutes becomes \(30 \, ^\circ\text{C}\), then the increase in the internal energy of the water is (Specific heat capacity of water \( = 4200 \, \text{J kg}^{-1} \text{ K}^{-1} \))

• A. 100 kJ
• B. 420 kJ
• C. 510 kJ
• D. 210 kJ

Hint:

For solids and liquids, the change in internal energy \( \Delta U \) due to a temperature change \( \Delta T \) (without phase change) is approximately \( \Delta U = mc\Delta T \), where \(m\) is mass and \(c\) is specific heat capacity. A change in temperature of \( X \, ^\circ\text{C} \) is equivalent to a change of \( X \, \text{K} \). 1 kJ = 1000 J. The information about heating time is not needed if we are calculating the change in internal energy based on temperature change.

Answer 1

The Correct Option is D

Mass of water \( m = 5 \) kg.
Initial temperature \( T_1 = 20 \, ^\circ\text{C} \).
Final temperature \( T_2 = 30 \, ^\circ\text{C} \).
Specific heat capacity of water \( c = 4200 \, \text{J kg}^{-1} \text{ K}^{-1} \).
The time of heating (10 minutes) is extra information if we assume all heat supplied goes into increasing internal energy and no phase change or work done.
For water (an incompressible liquid, approximately), the increase in internal energy \( \Delta U \) when its temperature changes by \( \Delta T \) is given by \( \Delta U = mc\Delta T \).
Change in temperature \( \Delta T = T_2 - T_1 = 30 \, ^\circ\text{C} - 20 \, ^\circ\text{C} = 10 \, ^\circ\text{C} \).
A temperature difference of \( 10 \, ^\circ\text{C} \) is equal to a temperature difference of \( 10 \, \text{K} \).
So, \( \Delta T = 10 \, \text{K} \).
Increase in internal energy: \[ \Delta U = mc\Delta T = (5 \text{ kg}) \times (4200 \, \text{J kg}^{-1} \text{ K}^{-1}) \times (10 \, \text{K}) \] \[ \Delta U = 5 \times 4200 \times 10 \, \text{J} \] \[ \Delta U = 50 \times 4200 \, \text{J} = 210000 \, \text{J} \] To convert Joules to kiloJoules (kJ), divide by 1000: \[ \Delta U = \frac{210000}{1000} \, \text{kJ} = 210 \, \text{kJ} \] This matches option (4).
The "closed vessel" implies no mass escapes.
If volume is constant and no work is done, then heat supplied = change in internal energy.