An object with a mass of #32 g# is dropped into #210 mL# of water at #0^@C#. If the object cools by #96 ^@C# and the water warms by #7 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Jane
Jan 26, 2018

Mass of #210 mL# of water is #210 cm^3 × 1 (gm)/(cm^3)# i.e #210 gm#

So, heat energy absorbed by that much water to increase its temperature by #7# degree C will be #(210×1×7)# Calorie or #14700# Calorie (using #H = m×s×d theta#)

And that amount of energy must have been released by that #32 gm# of object which has cooled by #96# degree C,whose value is #(32×s×96)# Calorie where #s# is the specific heat of that object.

So, equating both we get, #s = 4.78# Calorie #gm^-1 degree C ^-1 #

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