An object with a mass of #3 kg#, temperature of #173 ^oC#, and a specific heat of #23 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?

Question

1281 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-25T14:21:35

The water does not evaporate and the change in temperature is #=0.09^@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=173-T#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

#C_w=4.186kJkg^-1K^-1#

#C_o=0.023kJkg^-1K^-1#

#3*0.023*(173-T)=32*4.186*T#

#173-T=(32*4.186)/(3*0.023)*T#

#173-T=1941.3T#

#1942.3T=173#

#T=173/1942.3=0.09^@C#

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