An object with a mass of 7 kg, temperature of 163 ^oC, and a specific heat of 17 J/(kg*K) is dropped into a container with 32 L of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Mar 16, 2017

The water does not evaporate and the temperature change by 0.15ºC

Explanation:

The heat is transferred from the hot object to 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=163-T

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

C_w=4.186kJkg^-1K^-1

C_o=0.017kJkg^-1K^-1

7*0.017*(163-T)=32*4.186*T

163-T=(32*4.186)/(0.119)*T

163-T=1125.6T

1126.6T=163

T=163/1126.6=0.15ºC

The water does not evaporate