An object with a mass of 6 kg, temperature of 173 ^oC, and a specific heat of 29 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
Jun 6, 2017

The water does not evaporate and the change in temperature is =0.22ºC

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=173-T

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

C_w=4.186kJkg^-1K^-1

C_o=0.029kJkg^-1K^-1

6*0.029*(173-T)=32*4.186*T

173-T=(32*4.186)/(6*0.029)*T

173-T=769.8T

770.8T=173

T=173/770.8=0.22ºC

As T<100ºC, the water does not evaporate