An object with a mass of #1 kg#, temperature of #105 ^oC#, and a specific heat of #32 J/(kg*K)# is dropped into a container with #16 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1325 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-08T13:50:41

The water does not evaporate and the change in temperature is #=0.05^@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=105-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=0.032kJkg^-1K^-1#

The mass of the object is #m_0=1kg#

The mass of the water is #m_w=16kg#

#1*0.032*(105-T)=16*4.186*T#

#105-T=(16*4.186)/(1*0.032)*T#

#105-T=2093T#

#2094T=105#

#T=105/2094=0.05^@C#

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