(11%) Problem 1: Suppose you pour 0.0115 kg of 200°C water onto a 0.95-kg block
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(11%) Problem 1: Suppose you pour 0.0115 kg of 200°C water onto a 0.95-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) Otheexpertta.com > ? 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 Deductions Potential 0% 100% cosO cotan0 asin acos0 Submissions Attempts remaining: 20 sin) tan0( % per attempt) detailed view atan) acotan( cosh0 tanh cotanhO Degrees Radians Submit Hint Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 0% deduction per feedbackExplanation / Answer
final temperature is T then assuming no heat loss,
Q absorbed by ice = Q released by water
(0.95 x 2090 x (T - (-15))) = (0.0115 x 4186 x (20 - 0)) + (0.0115 x 334 x 10^3) + (0.011 x 2090 x (0 - T))
1985.5 T +29782.5 = 4803.78 - 24.035 T
T = - 12.4 deg C
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