A sample of an unknown metal mass 0.550 kg is heated in a pan of hot water until

Question

 A sample of an unknown metal mass 0.550 kg is heated in a  pan of  hot water until it is in thermal equilibrium with the water at a temperature of 75 degrees celcius . The sample is then removed from the water and placed into the inner cylinder ( aluminum) of a calorimeter that contains 0.500 kg of water at 15.5 degreee celcius . The mass of the inner cylinder is 0.100 kg . When the content of the calorimeter reach equilibrium the temperature inside is 18.8 degree celcius . Find the specific heat of the metal .

Explanation / Answer

the heat lost by the metal sample is equal to the heat gainedby the contents of the inner cylinder of a calorimeter therefore weget mw * sw * tw + mAl* sAl * tAl = m1 * s1 *(t - te) ---------(1) mw = 0.500 kg sw = 4186 J/kg/K tw = 75 oC = (75 + 273) K = 348 K mAl = 0.100 kg sAl = 900 J/kg/K tAl = 15.5 oC = (15.5 + 273) K = 288.5K m1 = 0.550 kg s1 is the specific heat of the metal (t - te) = (75 - 15.5) oC = 59.5oC = (59.5 + 273) K = 332.5 K solving equation (1) for s1 we have s1 = [mw * sw * tw+ mAl * sAl * tAl/m1 *(t - te)]

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