An object with a mass of 10 g is dropped into 240 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 8 ^@C, what is the specific heat of the material that the object is made of?

2 Answers
Jane
Feb 28, 2018

Here,the heat energy liberated by the object will be taken up by the water as a result of which,both will come to a thermal equilibrium.

So,heat energy liberated by the object is 10*s* 120 Calorie (using H= msd theta,where, m is the mass of the object, s is its specific heat and d theta is the change in temperature)

Now,heat energy absorbed by water is 240*1*1*8 calorie (where, mass of water = volume* density, and 1 ml =1 cm^3 and in CGS, density of water is 1 g/(cm^3) and specific heat of water is 1 C g^-1 @C ^-1)

So,equating both we get,

1200s =1920

so, s= 1.6 C g^-1 @C^-1

Narad T.
Feb 28, 2018

The specific heat is =6.76.7 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=120ºC

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=0.010kg

The mass of the water is m_w=0.24kg

0.010*C_o*120=0.24*4.186*8

C_o=(0.24*4.186*8)/(0.010*120)

=6.7 kJkg^-1K^-1

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