An object with a mass of 120 g is dropped into 720 mL of water at 0^@C. If the object cools by 24 ^@C and the water warms by 3 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
Shwetank Mauria
Sep 19, 2016

Specific heat of object is 0.75.

Explanation:

Let the specific heat of object be s.

As quantum of heal gained / lost is given by mxxsxx(t_2-t_1), where m is mass of object gaining / loosing heat and s is its specific heat and (t_2-t_1) is difference in temperature caused by heating / cooling,

heat lost by objece is 120xxsxx24

and heat gained by water is 720xx1xx3.

Now heat lost should be equal to heat gained

120xxsxx24=720xx1xx3

i.e. s=(720xx3)/(120xx24)=(6cancel720xxcancel3)/(cancel120xx8cancel24)=6/8=3/4=0.75

Hence specific heat of object is 0.75.

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