SATHEE: Heat and Thermodynamics 1 Question 10

Heat and Thermodynamics 1 Question 10

10. A water cooler of storage capacity 120 litres can cool water at a constant rate of $P$ watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly $3 k W$ of heat (thermal load). The temperature of water fed into the device cannot exceed $30^{\circ} C$ and the entire stored 120 litres of water is initially cooled to $10^{\circ} C$ . The entire system is thermally insulated. The minimum value of $P$ (in watts) for which the device can be operated for 3 hours is

(2016 Adv.)

(Specific heat of water is $4.2 k J k g^{- 1} K^{- 1}$ and the density of water is $1000 k g m^{- 3}$ )

(a) 1600

(b) 2067

(d) 3933

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Solution:

Heat generated in device in $3 h$

$= time \times power = 3 \times 3600 \times 3 \times 10^{3} = 324 \times 10^{5} J$

Heat used to heat water

$= m s Δ θ = 120 \times 1 \times 4.2 \times 10^{3} \times 20 J$

Heat absorbed by coolant

$\begin{aligned} = P t = 324 \times 10^{5} - 120 \times 1 \times 4.2 \times 10^{3} \times 20 J \\ P t & = (325 - 100.8) \times 10^{5} J \\ P & = \frac{223.2 \times 10^{8}}{3600} = 2067 W \end{aligned}$

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Heat and Thermodynamics 1 Question 10

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Heat and Thermodynamics 1 Question 10

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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