An object with a mass of 3 kg, temperature of 123 ^oC, and a specific heat of 14 J/(kg*K) is dropped into a container with 32 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 4, 2018

The water will not evaporate and the change in temperature is =0.04^@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=123-T

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

The specific heat of the object is C_o=0.014kJkg^-1K^-1

The mass of the object is m_0=3kg

The volume of water is V=32L

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

The mass of the water is m_w=rhoV=32kg

3*0.014*(123-T)=32*4.186*T

123-T=(32*4.186)/(3*0.014)*T

123-T=3189.3T

3190.3T=123

T=123/3190.3=0.04^@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 not very high and the specific heat is low.

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