An object with a mass of #12 kg#, temperature of #210 ^oC#, and a specific heat of #7 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
Jul 18, 2017

Answer:

#DeltaT=208.8^circC# The water will evaporate.

Explanation:

#210^circC~~483K#
#0^circC~~273K#
#32L=0.032m^3~=0.032kg#

#m_oC_oDeltaT_o=m_wC_wDeltaT_w#

#12*7*(483-T)=0.032*4.18*(273+T)#

#40572-84T=36.51648+0.13376T

#84.13376T=40535.4835#

#T=40535.4835/84.13376=481.798074K=208.798074^circC~~208.8^circC#

#0+208.8=208.8^circC#

As #208.8gt100# the water will evaporate.