An object with a mass of #12 kg#, temperature of #145 ^oC#, and a specific heat of #23 J/(kg*K)# is dropped into a container with #35 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1283 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-12T09:12:58

The water does not evaporate and the change in temperature is #=0.27^@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=145-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=0.023kJkg^-1K^-1#

The mass of the object is #m_0=12kg#

The mass of the water is #m_w=35kg#

#12*0.023*(145-T)=35*4.186*T#

#145-T=(35*4.186)/(12*0.023)*T#

#145-T=530.8T#

#531.8T=145#

#T=145/531.8=0.27^@C#

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