An object with a mass of #2 kg#, temperature of #315 ^oC#, and a specific heat of #12 (KJ)/(kg*K)# is dropped into a container with #37 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Sep 16, 2016

Answer:

The water does not evaporate. The final temperature of the water is:

#T=42^oC#

So the temperature change:

#ΔT=42^oC#

Explanation:

The total heat, if both remain in the same phase, is:

#Q_(t ot)=Q_1+Q_2#

Initial heat (before mixing)

Where #Q_1# is the heat of water and #Q_2# the heat of the object. Therefore:

#Q_1+Q_2=m_1*c_(p_1)*T_1+m_2*c_(p_2)*T_2#

Now we have to agree that:

  • The heat capacity of water is:

#c_(p_1)=1 (kcal)/(kg*K)=4,18(kJ)/(kg*K)#

  • The density of water is:

#ρ=1 (kg)/(lit)=>1lit=1kg-># so kg and liters are equal in water.

So we have:

#Q_1+Q_2=#

#=37kg*4,18(kJ)/(kg*K)*(0+273)K+2kg*12(kJ)/(kg*K)*(315+273)K#

#Q_1+Q_2=56334,18kJ#

Final heat (after mixing)

  • The final temperature of both the water and the object is common.

#T_1'=T_2'=T#

  • Also, the total heat is equal.

#Q_1'+Q_2'=Q_1+Q+2#

Therefore:

#Q_1+Q_2=m_1*c_(p_1)*T+m_2*c_(p_2)*T#

Use equation to find final temperature:

#Q_1+Q_2=T*(m_1*c_(p_1)+m_2*c_(p_2))#

#T=(Q_1+Q_2)/(m_1*c_(p_1)+m_2*c_(p_2))#

#T=(56334,18)/(37*4,18+2*12)(kJ)/(kg*(kJ)/(kg*K)#

#T=315^oK#

#T=315-273=42^oC#

Provided the pressure is atmospheric, the water did not evaporate, since its boiling point is #100^oC#. The final temperature is:

#T=42^oC#

So the temperature change:

#ΔT=|T_2-T_1|=|42-0|=42^oC#