ETA 16-17: Linear Expansion and Calorimetry Begin Date: 2/19/2018 2:00:00 AM --
ID: 3308786 • Letter: E
Question
ETA 16-17: Linear Expansion and Calorimetry Begin Date: 2/19/2018 2:00:00 AM -- Due Date: 2/27/2018 1:59:00 PM End Date: 5/11/2018 12:00:00 AM (14%) Problem 7: Suppose you pour 0.0125 kg of 200°C water onto a 1.35-kg block of ice, sitting in a large bowl, which is initially at -15.0°C. The latent heat of fusion for water is Lf 334 kJ/kg Specific heat (c) Substances Solids Aluminum Concrete Copper Glass Ice (average) 900 840 387 840 2090 kcal/kg.C 0.215 0.20 0.0924 0.20 0.50 Liquids Water 4186 1.000 Gases Steam (100°C) 1520 (2020)0.363(0.482) ©theexpertta.com p What is the final temperature of this system? You may assume that the water cools so rapidly that effects of the surroundings are negligible Grade Summary Potential Submissions Attempts remaining: 10 0% 100% sin tan() | | ( 78 9 coS cotan0 asinO acosO atan) acotan) sinh(0 % per attempt) detailed view cosh0tanh cotanh0 0 END Degrees O Radians DELCLEAR Submit Hint I give up! Hints: 3% deduction per hint. Hints remaining: 2 Feedback: 1% deduction per feedback.Explanation / Answer
from energy conservation,
heat relased by water = heat absorbed by ice
For Water: to decrease the temp to 0 deg C,
Q1 = m C deltat = (0.0125) (4.186 ) (20 - 0)
Q1 = 1.0465 kJ
to freeze the water" Q2 = m Lf = (0.0125)(334)
Q2 = 4.175 kJ
(1.35 x 2.05 (T - (-15))) = (0.0125 x 2.05(0 - T)) + 1.0465 + 4.175
2.7675 T + 41.5125 = - 0.025625 T + 1.0465 + 4.175
T = -13 deg C ....Ans
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.