An object with a mass of 32 g is dropped into 210 mL of water at 0^@C. If the object cools by 60 ^@C and the water warms by 7 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
Mar 21, 2018

The specific heat is =3.21 kJkg^-1K^-1

Explanation:

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

For the cold water, Delta T_w=7ºC

For the hot object DeltaT_o=60ºC

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=0.032kg

The mass of the water is m_w=0.21kg

0.032*C_o*60=0.21*4.186*7

C_0=(0.21*4.186*7)/(0.032*60)

=3.21 kJkg^-1K^-1