Class Management I Help Homework #10 Begin Date: 3/12/2018 12:00:00 AM-Due Date:
ID: 2036787 • Letter: C
Question
Class Management I Help Homework #10 Begin Date: 3/12/2018 12:00:00 AM-Due Date: 4/8/2018 11:59:00 PM End Date: 4/15/2018 12:00:00 AM (3%) Problem 34: You have mw-3.6 kg of water in an insulated container. You add m1 = 0.55 kg of ice at Tl--21°C to the 3 water and the mix reaches a final, equilibrium temperature of Tf= 19°C . The specific heats of ice and water are cl-2.10x10 J/(kg.°C) and cw 4.19x103 J/(kg.°C), respectively, and the latent heat of fusion for water is Lf 3.34x103 J/kg OACalculate the initial temperature of the water, in degrees Celsius Grade Summary Deductions Potential 090 100% sinO ?|( tan( ) acos) cosO Submissions cotanasin Attempts remaining: 5 % per attempt) detailed view atan acotansinh0 cosh tanh cotanh) 0 END Degrees Radians Submit Hint I give up! Hints: 290 deduction per hint. Hints remaining: 3 Feedback: 2% deduction per feedback. Submission History Answer Hints Feedback Totals Totals 0% 0% 0% 0%Explanation / Answer
Let Tw be the initial temperature of water.
For ice, Q is different from each process
(1) From T=-21 oC to T=0 :
Q = mc?T
where
m = mass (kg)
c = specific heat capacity (J/kg-oC)
T = temperature (?oC)
Q = ( 0.55 )( 2.10 x 103 )( 0 - (-21))
Q = 24.25 x 103 J
(2) Phase change of ice to water at constant T
Q = m?Lf
where Lf is the latent heat of fusion
Q = ( 0.55)( 3.34 x 105 )
Q = 183.7 x 103 J
(3) From T=0 to final Tf.
At this point, ice is now water:
Q = mc?Tf
Q = ( 0.55)( 4.19 x 103)(19 - 0)
Q = 43.8 x 103 J
For hot water initially at Tw, heat lost is
Q = mc?T
Q = ( 3.6)( 4.19 x 103 )( Tw -19)
Q = 15.08 x 103( Tw - 19)
Heat gained by ice = Heat lost by water
24.25 x 103 + 183.7 x 103 + 43.8 x 103 = 15.08 x 103( Tw - 19)
Tw = 35.7o C