An object with a mass of 6 kg, temperature of 50 ^oC, and a specific heat of 4 J/(kg*K) is dropped into a container with 24 L of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Narad T.
Mar 3, 2018

The water will not evaporate and the change in temperature is =0.012^@C

Explanation:

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

Let T= final temperature of the object and the water

For the cold water, Delta T_w=T-0=T

For the object DeltaT_o=50-T

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

The [
The specific heat of the object is C_o=0.004kJkg^-1K^-1

The mass of the object is m_0=6kg

The volume of water is V=24L

The density of water is rho=1kgL^-1

The mass of the water is m_w=rhoV=24kg

6*0.004*(50-T)=24*4.186*T

50-T=(24*4.186)/(6*0.004)*T

50-T=4186T

4187T=50

T=50/4187=0.012^@C

As the final temperature is T<100^@C, the water will not evaporate. We expect this result as the temperature of the object is <100^@C and the specific heat is low.

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