A 0.55-kg block of aluminum [c = 0.900 kJ/(kg middot K)] at 20.4 degree C is in

Question

A 0.55-kg block of aluminum [c = 0.900 kJ/(kg middot K)] at 20.4 degree C is in contact with a 0.55-kg block of silver [c = 0.235 kJ/(kg middot K)] at a temperature of 20.4 degree C. The system is completely isolated from the rest of the universe. Suppose heat flows from the aluminum into the silver until the temperature of the silver is 23.3 degree C. Part 1 out of 3 (a) Using the first law of thermodynamics, calculate the final temperature of the aluminum. Round intermediate calculations and your final answer to one decimal place. degree C

Explanation / Answer

From the principle of calorimetry we know that
heat lost by the aluminium = heat gain by the silver
Heat = mCdT
where m is mass , c is specific heat and dT is change in temperature
malCal(Tintial - Tfinal) = msilCSil(Tfinal - Tinital)
0.55*0.9*(20.4 - TFinal) = 0.55*0.235*(23.3 - 20.4)
Tfinal = 19.643 oC

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