An iron container has a mass of 200 g and contains 50 g of water @ 40Â°C. 50 g of ice @ -6Â°C are poured. Calculate the equilibrium temperature and describe the final composition.

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aristocles aristocles · Answer 1 · 2019-07-15T03:25:07+02:00

Answer:

final equilibrium temperature of the system is ZERO degree Celcius

Explanation:

Hear heat given by water + iron = heat absorbed by ice

so here first we will calculate the heat given by water + iron

[tex]Q_1 = m_1s_2\Delta T_1 + m_2 s_2 \Delta T_1[/tex]

[tex]Q_1 = (200)(0.450)(40 - T) + (50)(4.186)(40 - T)[/tex]

now the heat absorbed by ice so that it will melt and come to the final temperature

[tex]Q_2 = m s \Delta T + mL + m s_{water}\Delta T'[/tex]

[tex]Q_2 = 50(2.09)(0 + 6) + 50(335) + 50(4.186)(T - 0)[/tex]

now we will have

[tex]17377 + 209.3T = 3600 - 90T + 8372 - 209.3T[/tex]

[tex]17377 + 209.3T + 90T + 209.3T = 11972[/tex]

[tex]T = -10.6[/tex]

since it is coming out negative which is not possible so here the ice will not completely melt

so final equilibrium temperature of the system is ZERO degree Celcius