Answer:
[tex]40.4^{\circ}C[/tex]
Explanation:
First of all, we have to calculate the heat released by the steam whem it turns into liquid water. This is given by:
[tex]Q_s = m_s \lambda_v[/tex]
where
[tex]m_s = 10 g=0.010kg[/tex] is the mass of steam
[tex]\lambda_v=2.26\cdot 10^6 J/kg[/tex] is the latent heat of vaporization
Substituting,
[tex]Q_s=(0.010)(2.26\cdot 10^6)=22,600 J[/tex]
Now we find the heat absorbed by the ice when it turns into liquid water:
[tex]Q_i=m_i \lambda_f[/tex]
where
[tex]m_i = 50 g = 0.050 kg[/tex] is the mass of ice
[tex]\lambda_f=3.33\cdot 10^5 J/kg[/tex] is the latent heat of fusion
So,
[tex]Q_i=(0.050)(3.33\cdot 10^5)=16650 J[/tex]
Then, we call:
[tex]Q_1 = m_i C (T-0^{\circ}C)[/tex] the heat absorbed by the ice (now liquid water) till reaching the equilibrium temperature T, where
[tex]C=4186 J/kgC[/tex] is the specific heat capacity of water
[tex]Q_2=m_s C(100^{\circ}-T)[/tex] the heat released by the steam (now liquid water) till reaching the equilibrium temperature T
Since the total heat absorbed must be equal to the total heat released, we have:
[tex]Q_i+Q_1 = Q_2 + Q_s[/tex]
And now we can solve to find T:
[tex]Q_i + m_i C T = 100 m_s C - m_s C T + Q_s\\T=\frac{Q_s-Q_i-100m_s C}{(m_i+m_s)C}=40.4^{\circ}C[/tex]
The final temperature of the mixture of steam and ice is 40.36 ⁰C.
The given parameters:
The final temperature of the mixture of steam and ice is calculated as follows;
[tex]Q_{loss} = Q_{gain}\\\\mc(100 - T) +m L_v = mc(T - 0) + mL_f\\\\0.01\times 4186(100 - T) + 0.01(2.26 \times 10^6)= 0.05( 4186T) + 0.05(3.33\times 10^5)\\\\4186 -41.86T + 22,600 = 209.3T + 16,650\\\\10136= 251.16T\\\\T = \frac{10136}{251.16} \\\\T= 40.36 \ ^0C[/tex]
Thus, the final temperature of the mixture of steam and ice is 40.36 ⁰C.
Learn more about final temperature of mixtures here: https://brainly.com/question/22338034
