An object with a mass of #1 kg#, temperature of #320 ^oC#, and a specific heat of #14 J/(kg*K)# is dropped into a container with #16 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1429 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-14T09:41:49

The water does not evaporate and the change in temperature is #=0.07^@C#

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=320-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=0.014kJkg^-1K^-1#

The mass of the object is #m_0=1kg#

The volume of water is #V=16L#

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

The mass of the water is #m_w=rhoV=16kg#

#1*0.014*(320-T)=16*4.186*T#

#320-T=(16*4.186)/(1*0.014)*T#

#320-T=4784T#

#4785T=320#

#T=320/4785=0.07^@C#

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