An object with a mass of 120 g is dropped into 810 mL of water at 0^@C. If the object cools by 48 ^@C and the water warms by 8 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
Dec 31, 2017

The specific heat is =4.71 kJkg^-1K^-1

Explanation:

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

For the cold water, Delta T_w=8ºC

For the object DeltaT_o=48ºC

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=0.12kg

The mass of the water is m_w=0.81kg

0.12*C_o*48=0.81*4.186*8

C_o=(0.81*4.186*8)/(0.12*48)

=4.71 kJkg^-1K^-1