An object with a mass of 50 g50g is dropped into 160 mL160mL of water at 0^@C0C. If the object cools by 40 ^@C40C and the water warms by 8 ^@C8C, what is the specific heat of the material that the object is made of?

1 Answer
Dec 14, 2017

The specific heat is =2.68 kJkg^-1K^-1=2.68kJkg1K1

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=40ºC

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=0.05kg

The mass of the water is m_w=0.16kg

0.05*C_o*40=0.16*4.186*8

C_o=(0.16*4.186*8)/(0.05*40)

=2.68 kJkg^-1K^-1