An object with a mass of 3 kg, temperature of 234 ^oC, and a specific heat of 37 (KJ)/(kg*K) is dropped into a container with 15 L of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Mar 12, 2017

The water evaporates

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=234-T

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

C_w=4.186kJkg^-1K^-1

C_o=37kJkg^-1K^-1

3*37*(234-T)=15*4.186*T

234-T=(15*4.186)/(111)*T

234-T=0.566T

1.566T=234

T=234/1.566=149.5ºC

The water evaporates