An object with a mass of #16 kg#, temperature of #270 ^oC#, and a specific heat of #5 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
Dec 29, 2017

Answer:

The water does not evaporate and the change in temperature is #=0.26^@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=270-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=0.005kJkg^-1K^-1#

The mass of the object is #m_0=16kg#

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

#16*0.005*(270-T)=32*4.186*T#

#270-T=(32*4.186)/(16*0.005)*T#

#270-T=1674.4T#

#1675.4T=270#

#T=270/1675.4=0.16^@C#

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