(8%) Problem 4: The same heat transfer into identical masses of different Specif

Question

(8%) Problem 4: The same heat transfer into identical masses of different Specific heat (c) Substances substances produces different temperature changes, due to differences in the heat Solids J/kg C kcal/kg. capacity of the various materials. 0.215 Aluminum 900 840 020 Concrete 387 Copper 0.0924 Glass 840 0.20 0.83 Human Body (37C) 3500 452 0,108 Iron, steel Liquids Water 4136 1.000 130 Mercury 0.0333 Otheexpertta.com A 25% Part (a) Calculate the final temperature, in degrees Celsius, when 0.95 kcal of heat transfers into I.05 kg of water, originally at 20.0°C Grade Summary Deductions 100% Potential sino cos0 tan0 Submissions 8 9 HOME Attempts remaining S cotano asino acos0 r attempt) detailed view atan0 acotano sinh0 cosh0 tanh0 cotanh Degrees Radians BACKSPACE Feedback Submit Hint I give up Hints 1% deduction per hint. Hints remaining Feedback 1% eduction per feedback. A 25% Part (b) Calculate the final temperature, in degrees Celsius, when 0.95 kcal of heat transfers into 1.05 kg of concrete, originally at 20.0°C A 25% Part (c) Calculate the final temperature, in degress Celsius, when 0.95 kcal of heat transfers into 1.05 kg of the steel, originally at 20.0°C A 25% Part (do Calculate the final temperature, in degrees Celsius, when 0.95 kcal of heat transfers into 1.05 kg of the mercury, originally at 20.0°C

Explanation / Answer

amount of heat transferred Q = m*S*dT

Q = m*S*(t2-t1)

part(a)

for water specific heat s = 1

0.95 = 1.05*1*(t2-20)

Tw = 20.9 degrees

part(b)

for concrete specifice heat s = 0.2

0.95 = 1.05*0.2*(t2-20)

t2 = 24.5 degrees

part(c)

for steel specific heat s = 0.108

0.95 = 1.05*0.108*(t2-20)

t2 = 28.4 degrees

part(d)

for mercury specific heat s = 0.0333

0.95 = 1.05*0.0333*(t2-20)

t2 = 47.2 degrees

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