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 #36 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Andrea B.
Jun 19, 2017

water's temperature change = 0,16°C

Explanation:

The object decrease its temperature toward a final temperature Tf giving an amount of heat that is:
# Q = m cp (Ti-Tf)= 4 kg 25 J/(kg K) (Ti-Tf) = 100 (J)/K(Ti-Tf) #

The water increases its temperature acquiring an egual amount of heat:# Q= m cp (Tf-T°)= 36 kg 4186 J/(kg K) (Tf-T°)= 150 000(J)/K(Tf-T°)#
Equalizing the two quantities of heat you have:
#100 (J)/K(240°C-Tf) =150 000(J)/K(Tf-0°C)#
Simplifying
#2(240°C-Tf) =3000(Tf-0°C)#
and resolving
480 - 2Tf = 3000 Tf
3002 Tf = 480
Tf = 0,16 °C
note that: 36 L of water are 36 kg; specific heat of water is (4186 J)/kg K); the specific heat of the object is really very small

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