Wednesday, September 28, 2016

A thermos bottle contains 0.150 kg of water at 4.1 degrees celsius. When 9 x 10^-2 kg of a metal, initially at 96.2 degrees celsius, is put into...

The amount of thermal energy gained or lost is calculated as:


thermal energy gained or lost = mass of substance x specific heat x change in temperature.


In this case, the temperature of water has increased and it has gained some thermal energy. This amount of thermal energy can be calculated as:


Thermal energy gained by water = mass of water x specific heat of water x (final temperature - initial temperature)


= 0.15 kg x 4186 J/kg/K x (21.7 - 4.1) C = 11,051.04 J


The gain in thermal energy of the water is due to thermal energy loss by the metal, which can be written as:


thermal energy lost by metal = 9 x 10^-2 x C x (96.2 - 21.7)


where C is the specific heat of the metal in J/kg/K.


Assuming no heat loss,


thermal energy lost by metal = thermal energy gained by water


or, 9 x 10^-2 x C x (96.2 - 21.7) = 11,051.04


solving the equation, we get, C = 1648.2 J/kg/K or 1.648 J/g/K

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