An object with a mass of 1 kg1kg, temperature of 120 ^oC120oC, and a specific heat of 32 J/(kg*K)32JkgK is dropped into a container with 12 L 12L of water at 0^oC 0oC. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Apr 18, 2018

The water will not evaporate and the change in temperature is =0.08^@C=0.08C

Explanation:

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

Let T=T= final temperature of the object and the water

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

For the object DeltaT_o=120-T

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

The specific heat of the object is C_o=0.032kJkg^-1K^-1

The mass of the object is m_0=1kg

The volume of water is V=12L

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

The mass of the water is m_w=rhoV=12kg

1*0.032*(120-T)=12*4.186*T

120-T=(12*4.186)/(1*0.032)*T

120-T=1569.8T

1570.8T=120

T=120/1570.8=0.08^@C

As the final temperature is T<100^@C, the water will not evaporate.