An object with a mass of #5 kg#, temperature of #163 ^oC#, and a specific heat of #14 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
Feb 27, 2017

Answer:

The water does not evaporate and the temperature of the water changec by #0.09º#C

Explanation:

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=163-T#

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

#C_w=4186Jkg^-1K^-1#

#C_o=14Jkg^-1K^-1#

#m_0 C_o*(163-T) = m_w* 4186 *T#

#5*14*(163-T)=32*4186*T#

#160-T=(32*4186)/(70)*T#

#163-T=1913.6T#

#1914.6T=163#

#T=163/1914.6=0.09ºC#

The water does not evaporate