Question

In a perfectly insulated container of negligible mass, 4.00 × 10−2 kg of steam at 100◦C and atmospheric pressure is added to 0.200 kg of water at 50.0◦C.

A) If no heat is lost to the surroundings, what is the final temperature of the system? B) At the final temperature, how many kilograms are there of steam and how many of liquid water?

Answer 1

Following are the solution to this question:

Step-by-step explanation:

Given value:

`m_s= 4.00 * 10^(-2) \ kg \\\\L_v=2256 * 10^(3) \ (J)/(kg)\\\\m_w= 0.2 \ kg\\\\\Delta T= 50^(\circ)`

In point A:

`Q_(Steam)=m_s \ L_v`

`=0.04 * 2256 * 10^(3)\\\\=9.02 * 10^4 \ J`

`Q_(water)= m_w \ c_w \ \Delta T\\\\`

`=0.2 * 4190 * 50\\\\=4.19 * 10^4 \ J`

`Q_(steam)> Q_(water)`, that's why the final temperature is
`= 100^(\circ)`

In point B:

`\to \Delta m_s L_v=m_w\ c_w \Delta T\\\\\to \Delta m_s * 2256* 10^3= 0.2 * 4190 * 50\\\\\to \Delta m_s= 1.86 * 10^(-2) \ kg\\\\\to m_s = 2.14 * 10^(-2) \ kg\\\\\to liquid \ left = 0.2+ 2.14 * 10^(-2) = 2.34 * 10^(-2) \\`