Explanation Write the general expression for the energy balance equation. E in − E out = Δ E system (I) Here, the total energy entering the system is E in , the total energy leaving the system is E out , and the change in the total energy of the system is Δ E system . Simplify Equation (I) and write energy balance for vacant room. Δ t × W electric heater − Q loss = Δ U = m Δ u = m c ν Δ T (II) Here, the heat is lost from the room is Q loss , the work to be done by the baseboard electric resistance heater is W electric heater , the change in the internal energy is Δ U , the mass of air in the room is m , The change in specific internal energy of a water is Δ u ,the specific heat of constant volume for air is c ν , the change in temperature of vacant room is Δ T , and the change of time of vacant room is Δ t . In this case, the vacant room is tightly sealed, so there is no heat loss in the vacant room. Substitute 0 for Q loss in Equation (II). Δ t × W electric heater = m c ν Δ T (III) Conclusion: Substitute 75 kg for m , 15 min for Δ t , 2 kW for W ˙ electric heater , 0 kJ , and 0

8th Edition

ISBN: 9780073398174

Author: Yunus A. Cengel Dr., Michael A. Boles

Publisher: McGraw-Hill Education

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Chapter 4.5, Problem 148FEP

To determine

The temperature rise of air at the end of 15 min is:

(a) 8.5°C.

(b) 12.4°C.

(c) 24.0°C.

(d) 33.4°C.

(e) 54.8°C.

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Chapter 4.5, Problem 147FEP

Chapter 4.5, Problem 149FEP

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The temperature rise of air at the end of 15 min is: (a) 8.5 ° C . (b) 12.4 ° C . (c) 24.0 ° C . (d) 33.4 ° C . (e) 54.8 ° C .

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The temperature rise of air at the end of 15 min is: (a) 8.5°C. (b) 12.4°C. (c) 24.0°C. (d) 33.4°C. (e) 54.8°C.

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Chapter 4 Solutions

The temperature rise of air at the end of 15 min is:

(a) 8.5°C.

(b) 12.4°C.

(c) 24.0°C.

(d) 33.4°C.

(e) 54.8°C.