An object with a mass of #6 kg#, temperature of #173 ^oC#, and a specific heat of #19 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

1264 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-04T07:30:19

The water does nor evaporate and the change in temperature is #=0.15^@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)#

The specific heat of water is #C_w=4.186kJkg^-1K^-1#

The specific heat of the object is #C_o=0.019kJkg^-1K^-1#

The mass of the object is #m_0=6kg#

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

#6*0.019*(173-T)=32*4.186*T#

#173-T=(32*4.186)/(6*0.019)*T#

#173-T=1175T#

#1176T=173#

#T=173/1176=0.15^@C#

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