4:42 PM OO AT&T; LTE 43% usi38ok.theexpertta.com C at Chapters 14 and 15 Begin D

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4:42 PM OO AT&T; LTE 43% usi38ok.theexpertta.com C at Chapters 14 and 15 Begin Date 4/27/2017 12:00:00 AM Due Date: 5/6/2017 ll:D te: 5/6/2017 11:00:00 PM Problem An unknown material, of degrees Cis added to a Dewer Can insulated ontainer which contains m2 1.3 kg of water at T 2 degrees Water has a specific heat ofc. 4186 UokgK). Ater the ystem contesto equilibrium the final temperature is T 31 degrees C. a 50% Part (a) Input an expression for the specific heat off the unknown material. a 50% Part (b) What is the specific heat in J/kg K)? ODegrees Radians All content 2017 Expert TA, LLC

Explanation / Answer

Let Cm be the specific heat of material.

Heat lost by the material = Heat gained by the water

or     m1*Cm*( T1 - T ) = m2*Cw*( T - T2 )

a) Cm = [m2*Cw*( T - T2 )] / [m1*( T1 - T )]

b) Cm = 1.3*4186*(31-23)/(0.44*(81 - 31) = 1978.84 J/(kg-K)

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