Question

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

Answer 1

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

$Q_1 = m_1s_2\Delta T_1 + m_2 s_2 \Delta T_1$

$Q_1 = (200)(0.450)(40 - T) + (50)(4.186)(40 - T)$

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

$Q_2 = m s \Delta T + mL + m s_{water}\Delta T'$

$Q_2 = 50(2.09)(0 + 6) + 50(335) + 50(4.186)(T - 0)$

now we will have

$17377 + 209.3T = 3600 - 90T + 8372 - 209.3T$

$17377 + 209.3T + 90T + 209.3T = 11972$

$T = -10.6$

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