An object with a mass of 32 g is dropped into 480 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 4 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
Shwetank Mauria
Sep 23, 2016

Specific heat of object is 0.50.

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 object is 32xxsxx120

and heat gained by water is 480xx1xx4.

Now heat lost should be equal to heat gained

32xxsxx120=480xx1xx4

i.e. s=(480xx4)/(32xx120)=(cancel4xxcancel120xxcancel4)/(cancel4xxcancel4xx2xxcancel120)=1/3=0.50

Hence specific heat of object is 0.50.

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