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

Answer:

Following are the solution to this question:

Explanation:

Given value:

$m_s= 4.00 \times 10^{-2} \ kg \\\\L_v=2256 \times 10^{3} \ \frac{J}{kg}\\\\m_w= 0.2 \ kg\\\\\Delta T= 50^{\circ}$

In point A:

$Q_{Steam}=m_s \ L_v$

            $=0.04 \times 2256 \times 10^{3}\\\\=9.02 \times 10^4 \ J$

$Q_{water}= m_w \ c_w \ \Delta T$

           $=0.2 \times 4190 \times 50\\\\=4.19 \times 10^4 \ J$

$Q_{steam}> Q_{water}$, that's why the final temperature is $= 100^{\circ}$

In point B:

$\to \Delta m_s L_v=m_w\ c_w \Delta T\\\\\to \Delta m_s \times 2256\times 10^3= 0.2 \times 4190 \times 50\\\\\to \Delta m_s= 1.86 \times 10^{-2} \ kg\\\\\to m_s = 2.14 \times 10^{-2} \ kg\\\\\to liquid \ left = 0.2+ 2.14 \times 10^{-2} = 2.34 \times 10^{-2}$