An object with a mass of #4 kg#, temperature of #240 ^oC#, and a specific heat of #25 J/(kg*K)# is dropped into a container with #24 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
May 21, 2017

Answer:

No. Only 0.239’C

Explanation:

The specific heat of water is #4.184 J/(g*’C)#. The water mass is 24000g (1000g/L). The energy required to heat it is therefore
#24000g * 4.184 (J/(g*’C)) * (T_2 - 0)’C #

The object supplies #4kg * 25(J/(kg*’C)) * (240 – T_2)’C#

Equating these two and solving for #T_2# will show us whether the water will evaporate or not (#T_2 > 100’C#).

#24000g * 4.184 (J/(g*’C)) * (T_2 - 0)’C # = #4000g * 0.025(J/(g*’C)) * (240 – T_2)’C#

#100416 * (T_2 - 0)’C # = #100 * (240 – T_2)’C#

#100416 * T_2’C # = #24000 -100 * T_2’C#

#100516 * T_2’C # = #24000’C# ; #T_2’C # = 0.239’C#

The water is barely heated. Its large heat capacity is one reason that it is such a good industrial process coolant.