Explanation Write the change in pressure using ideal gas relation. c p − c v = − T ( ∂ v ∂ T ) P 2 ( ∂ P ∂ v ) T (I) Here, total temperature is T , partial derivative of volume v with respect to temperature T is ( ∂ v ∂ T ) P and partial derivative of pressure P with respect to volume v is ( ∂ P ∂ v ) T . Rewrite Equation (I) in difference form at specified state. c p − c v = − T ( Δ v Δ T ) P = 15 MPa 2 ( Δ P Δ v ) T = 80 ° C (II) Here, total temperature is T , change in temperature at 80 ° C is Δ T , change in pressure at 15 MPa is Δ P , change in volume is Δ v , specific heat at constant pressure is c p and specific heat at constant volume is c v . Conclusion: Substitute v 1 − v 2 for Δ v , T 1 − T 2 for Δ T and P 1 − P 2 for Δ P in Equation (II). c p − c v = − T ( v 1 − v 2 T 1 − T 2 ) P = 15 MPa 2 ( P 1 − P 2 v 1 − v 2 ) T = 80 ° C (IV) Here, initial volume is v 1 , final volume is v 2 , initial temperature is T 1 , final temperature is T 2 , initial pressure is P 1 and final pressure is P 2 . Refer Table A-4, “Saturated water temperature table”, obtain the volume of saturated liquid v f at temperature of 100 ° C and 60 ° C . ( v 1 ) 100 ° C = 0.0010361 m 3 / kg ( v 2 ) 60 ° C = 0

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ISBN: 9780073398174

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

Publisher: McGraw-Hill Education

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A substance with fixed chemical composition throughout is called a pure substance like water, air, and nitrogen. A pure substance does not have to be a single element or compound. A mixture of two or more phases of a pure substance is also a pure substan…

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Chapter 12.6, Problem 33P

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The specific heat difference for (cp−cv) liquid water.

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Chapter 12.6, Problem 32P

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The specific heat difference for ( c p − c v ) liquid water.

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