An object with a mass of 18 g is dropped into 360 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 2 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
Jun 18, 2017

The specific heat is =1.4kJkg^-1K^-1

Explanation:

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

For the cold water, Delta T_w=2ºC

For the object DeltaT_o=120ºC

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

Mass of the object is m_0=0.018kg

Mass of the water is m_w=0.36kg

0.018*C_o*120=0.36*4.186*2

C_o=(0.36*4.186*2)/(0.018*120)

=1.4kJkg^-1K^-1