# 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?

May 21, 2017

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.