An object with a mass of 100 g is dropped into 400 mL of water at 0^@C. If the object cools by 20 ^@C and the water warms by 4 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
Apr 15, 2017

The specific heat is =3.35kJkg^-1K^-1

Explanation:

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

For the cold water, Delta T_w=4ºC

For the object DeltaT_o=20ºC

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

C_w=4.186kJkg^-1K^-1

Let, C_o be the specific heat of the object

0.1*C_o*20=0.4*4.186*4

C_o=(0.4*4.186*4)/(0.1*20)

=3.35kJkg^-1K^-1