Class Management Help Homework #10 Begin Date: 3/12/2018 12:00:00 AM-Due Date: 4
ID: 2036782 • Letter: C
Question
Class Management 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 32: A 0.035-kg ice cube at-30.0°C is placed in 0.39 kg of 35.0°C water in a very well-insulated container. The latent heat of fusion for water is Lf- 79.8 kcal/kg Specific heat (c) Substances Solids J/kg C kcal/kg.C Aluminum Concrete 900 840 387 840 2090 0.215 0.20 0.0924 0.20 0.50 Glass Ice (average) Water Steam (100°C) Liquids 4186 1.000 Gases 1520 (2020) 0.363(0.482) Otheexpertta.com What is the final temperature of the water, in degrees Celsius? Grade Summary Deductions Potential 37.32 5% 95% sin() cos) ?|( 7 89 HOME tan( ) acos) Submissions Attempts remaining: 4 (5% per attempt) detailed view cotanasin0 atan acotan) sinh(0 cosh th cotanh) Degrees Radians END 500 () BACKSPACE DELI CLEAR Submit Hint I give up! Hints: 2% deduction per hint. Hints remain?ng Feedback: 2% deduction per feedback. Submission History Answer Hints Feedback Totals Tf 37.32 5% 090 090 | 596 Totals 5% 0% 090 | 596Explanation / Answer
given
m_Ice = 0.035 kg,
m_water = 0.39 kg at 35 C
Lf = K cal/kg
C_water = 1 kCal/(kg C)
let T is the final equilibrium temperature.
Apply heat lost by water = heat gained by Ice
m_water*C_water*(35 - T) = m_Ice*Lf + m_Ice*C_Water*(T - 0 )
0.39*1*(35 - T) = 0.035*79.8 + 0.035*1*(T - 0)
==> T = 25.5 degrees celsius <<<<<<<-----------Answer