An object with a mass of 1 kg1kg, temperature of 150 ^oC150oC, and a specific heat of 24 J/(kg*K)24JkgK is dropped into a container with 27 L 27L of water at 0^oC 0oC. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Dec 16, 2017

The water does not evaporate and the change in temperature is =0.036^@C=0.036C

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=150-T

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

The specific heat of water is C_w=4.186kJkg^-1K^-1

The specific heat of the object is C_o=0.024kJkg^-1K^-1

The mass of the object is m_0=1kg

The mass of the water is m_w=27kg

1*0.024*(150-T)=27*4.186*T

150-T=(27*4.186)/(1*0.024)*T

150-T=4186T

4187T=150

T=150/4187=0.036^@C

As T<100^@C, the water does not evaporate