An object with a mass of #3 kg#, temperature of #360 ^oC#, and a specific heat of #24 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?

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Answer 1 · 2018-03-03T14:08:35

The water does not evaporate and the change in temperature is #=0.19^@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=360-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=0.024kJkg^-1K^-1#

The mass of the object is #m_0=3kg#

The volume of water is #V=32L#

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

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

#3*0.024*(360-T)=32*4.186*T#

#360-T=(32*4.186)/(3*0.024)*T#

#360-T=1860.4T#

#1861.4T=360#

#T=360/1861.4=0.19^@C#

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