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Measuring the change in temperature versus time upon addition of a hot metal specimen to 10.02...

Measuring the change in temperature versus time upon addition of a hot metal specimen to 10.02 ml of water with an initial temperature of 26.63 degrees C led to a plot with a line with the equation y= -0.121 x + 36.2. If the heat absorbed by the water is equal to the heat lost by the metal, determine the specific heat of the metal with a mass of 10.69g to two decimal places.

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Solution :- Given that: - Y -0.191% +36.2 Y-0.121 X+36-2. hd Metal Specimen =10.02m! Tempelature - 26.63. Mass -10.699. changSe, Given mw.cw (tv-TW) = mp ce (TR-Tew) 10.02 x1 x (36-397– 26.63) = 10-69xCpx (36-2-36-24) 10.02.X 9.599 = 10.69 XCRC-0.029

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