An object with a mass of 2 kg2kg, temperature of 244 ^oC244oC, and a specific heat of 33 (KJ)/(kg*K)33KJkgK is dropped into a container with 16 L 16L of water at 0^oC 0oC. Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Dec 17, 2017

The water will evaporate.

Explanation:

The heat is transferred from the hot object to the cold water.

Let T=T= final temperature of the object and the water

For the cold water, Delta T_w=T-0=T

For the object DeltaT_o=244-T

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

The specific heat of water is C_w=4.186kJkg^-1K^-1

The specific heat of the object is C_o=33kJkg^-1K^-1

The mass of the object is m_0=2kg

The mass of the water is m_w=16kg

2*33*(244-T)=16*4.186*T

244-T=(16*4.186)/(2*33)*T

244-T=1.015T

2.015T=244

T=244/2.015=121.1^@C

As T>100^@C, the water will evaporate