An object with a mass of #3 kg#, temperature of #155 ^oC#, and a specific heat of #37 (KJ)/(kg*K)# is dropped into a container with #15 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Question

1221 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-19T06:46:36

The final temperature of the water is #=99#ºC

The heat transferred from the hot object, is equal to the heat absorbed by 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=155-T#

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

#C_w=4.186KJkg^-1K^-1#

#C_o=37kJkg^-1K^-1#

#m_0 C_o*(155-T) = m_w* 4.186 *T#

#3*37*(155-T)=15*4.186*T#

#155-T=(15*4.186)/(3*37)*T#

#155-T=0.566T#

#1.566T=155#

#T=155/1.566=99ºC#