Self-Assessment Quiz #2 Begin Date: 10/21/2017 12:01:00 AM .. Due Date: 1 1 29 2

Question

Self-Assessment Quiz #2 Begin Date: 10/21/2017 12:01:00 AM .. Due Date: 1 1 29 2017 I 1:39:00 PM End Date: 11.29 20171 1:59:00 PM (20% ) Problem 5: A thermos contains m1 = 0.89 kg of tea at T,-33° C. Ice (M7=0.085 kg. T2 =0°C)is added to it. The heat capacity of both water and tea is c 4186 JkgK), and the latent heat of fusion for water is Ly- 335 x 10 Jkg. > 50% Part (a) Input an expression for the final temperature after the ice has melted and the system has reached thermal equilibrium. Grade Summan Amempts remaiti ed Feedhack I give up Hints:3for a 3 deduction Hints remaining 0 Feedback: 2 dedaction per feedhack There are three parts of the syslem- the lea, the lioe, and the wascr (once The sum of the heat change for each of them must be nero since the What is the initial the a closed. temperature for the water? Does the ioe change its b What is the final temperature in Kelvin? 888 a 3 5 8 9

Explanation / Answer

here,

m1 = 0.89 kg

m2 = 0.085 kg

let the final temprature be Tf

heat lost by water = heat gained by ice

m1 * Cw * ( T1 - Tf ) = m2 * ( Lf + Cw * ( Tf - 0))

((m1/m2) * Cw * ( T1 ) - Lf) = (m1/m2) * Cw * Tf + Cw * Tf

Tf = ((m1/m2) * Cw * ( T1 ) - Lf)/(( m1/m2) * Cw + Cw)

b)

Tf = ((0.89/0.085) * 4186 * ( 33 ) - 335000)/(( 0.89/0.085) * 4186 + 4186)

Tf= 23.1 degree C = 296.25 K

the final temprature is 296.25 K

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