(5%) Problem 21: A 0.24-kg aluminum bowl holding 0.55 kg of soup at 25.0°C is pl

Question

(5%) Problem 21: A 0.24-kg aluminum bowl holding 0.55 kg of soup at 25.0°C is placed in a freezer Specific heat (c) Substances Solids J/kg C kcal/kg C Aluminum Concrete Copper Glass Ice (average 0.215 0.20 900 840 387 840 2090 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, in degrees Celsius, if 377 kJ of energy is transferred from the bowl and soup, assuming the soup's thermal properties are the same as that of water? Grade Summary Deductions Potential Tf-- 4.52 10% 90% sinO tan 7 89 HOME cos cotanasin acosO atan acotan snh cosh tanh cotahO Submissions Attempts remaining: 3 (5% per attempt) detailed view 590 5% END Degrees Radians O BACKSPACE EL CLEAR Submit Hint I give up! Hints: 290 deduction per hint. Hints r emaining:1 Feedback: 2% deduction per feedback. Submission History Answer Hints Feedback Totals 5% 090 096 590 Tf-41.43 Tf=-4.52 5% 0% 0% 5%

Explanation / Answer

here,

mass of aluminum , m1 = 0.24 kg

mass of soup , m2 = 0.55 kg

let the final temprature be Tf

heat energy = m1 * Ca * ( 25 - Tf) + m2 * Cw * ( 25 - Tf) + m2 * Lf + m2 * Ci * ( 0 - Tf)

377 * 10^3 = 0.24 * 900 * ( 25 - Tf) + 0.55 * 4186 * ( 25 - 0) + 0.55 * 334000 + 0.55 * 2090 * (0 - Tf)

solving for Tf

Tf = - 95.4 degree C

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