Concepts Of Physics MCQ Edition [Volume 2]
- Exam
- Concepts Of Physics MCQ Edition [Volume 2]
- Subject
- Physics
- Chapter
- Calorimetry
Concepts Of Physics MCQ Edition [Volume 2] · Physics · Calorimetry
A traditional method of cooling drinking water is to keep it in a...
A traditional method of cooling drinking water is to keep it in a pitcher having porous walls. Water reaches the outer surface very slowly and evaporates. Most of the energy required for evaporation is extracted from the water itself, thereby cooling it down. Assume a pitcher holds of water and of water seeps out per second. Assuming no backward heat transfer from the atmosphere to the water, determine the time in which the temperature drops by 5^{}^{{}^{\circ}} C. The specific heat capacity of water is 4200 \text{ J kg}^{-1} ^{}^{{}^{\circ}} C^{-1} and the latent heat of vaporization of water is .
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
Let be the time required for the temperature to drop by 5^{}^{{}^{\circ}} C.
The rate of heat loss due to evaporation is given by:
Substituting the given values:
The total heat to be extracted from the water to cool it by 5^{}^{{}^{\circ}} C is:
Q = 10 \text{ kg} \times 4200 \text{ J kg}^{-1} ^{}^{{}^{\circ}} C^{-1} \times 5^{}^{{}^{\circ}} C = 210000 \text{ J}
Equating the total heat extracted to the heat lost over time :
Converting the time into minutes:
The rate of heat loss due to evaporation is given by:
Substituting the given values:
The total heat to be extracted from the water to cool it by 5^{}^{{}^{\circ}} C is:
Q = 10 \text{ kg} \times 4200 \text{ J kg}^{-1} ^{}^{{}^{\circ}} C^{-1} \times 5^{}^{{}^{\circ}} C = 210000 \text{ J}
Equating the total heat extracted to the heat lost over time :
Converting the time into minutes: