A solid ball of 2cm radium, made of steel (c = 486J/kg*K), isplaced in a flame o

Question

A solid ball of 2cm radium, made of steel (c = 486J/kg*K), isplaced in a flame of unknown temperature for a long time. The ballis then dropped into a copper cup of mass 250g containing 500gwater. The cup and water are initially at 20 degrees Celsius. Aftersome time has passed, the temperature of the system is found to be22.5 degrees Celsius. Treat the system as a thermally isolatedsystem.
a) Determine the temperature of the flame; b) How much heat was stored within the ball?
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a) Determine the temperature of the flame; b) How much heat was stored within the ball?
Please show all work!I promise to rate you as LIFESAVER if youranswer is correct!!!

Explanation / Answer

The amount of heat stored in the ball is found by the amountof heat released to the system of the copper and water. Youneed the specific heat of water and of copper water=  4.179 J/g*C copper= 0.385 J/g*C Since the whole system was raised in temperature then you needto times the temperature difference by both values and by thierweight to get the total heat. 4.179*2.5*500=5223.75 J .385*2.5*250=240.625 J Add them to get the total heat stored in the ball 5223.75 J +240.625 J=5464.375 J Then to find the temperature of the flame you need to know theweight of the ball. The density fo steel is 7.85 g/cm3 The volume is found by 4/3 pi r3. volume=4/3 pi 2^3= 33.5 cm^3 33.5*7.85=263 g= .263 kg. Tehn use the specific heat to find the temperature of theflame. 5464.375/486=11.24 kg *K T=11.24/.263=42.75 K. That is the temperature above 22.5 degreescelcius. 273.15+22.5+42.75=338.4 K=65.25 C For part b you can either use the answer 5464.375 J or you cancalculate the heat still in the ball from the 2.5 degree differenceadn add that on top. The volume is found by 4/3 pi r3. volume=4/3 pi 2^3= 33.5 cm^3 33.5*7.85=263 g= .263 kg. Tehn use the specific heat to find the temperature of theflame. 5464.375/486=11.24 kg *K T=11.24/.263=42.75 K. That is the temperature above 22.5 degreescelcius. 273.15+22.5+42.75=338.4 K=65.25 C For part b you can either use the answer 5464.375 J or you cancalculate the heat still in the ball from the 2.5 degree differenceadn add that on top.

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