An object with a mass of 20 g is dropped into 750 mL of water at 0^@C. If the object cools by 90 ^@C and the water warms by 18 ^@C, what is the specific heat of the material that the object is made of?

2 Answers
Jane
Mar 4, 2018

suppose,the specific heat of the object is s Cg^-1^@C^-1

So,during this process of achieving thermal equilibrium,the heat energy released by the object was taken by the water.

So,we can write,

20*s*90 = 750*1*18 (using, H=msd theta,where, d theta is the change in temperature,and specific heat for water is 1 CGS unit and weight of 750mL of water is 750g)

So,we get, s=7.5 Cg^-1^@C^-1

Narad T.
Mar 4, 2018

The specific heat is =31.4 kJkg^-1K^-1

Explanation:

The heat is transferred from the hot object to the cold water.

For the cold water, Delta T_w=18ºC

For the hot object DeltaT_o=90ºC

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

The specific heat of water C_w=4.186kJkg^-1K^-1

Let, C_o be the specific heat of the object

The mass of the object is m_o=0.020kg

The mass of the water is m_w=0.75kg

0.020*C_o*90=0.75*4.186*18

C_0=(0.75*4.186*18)/(0.020*90)

=31.4 kJkg^-1K^-1

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