Question

1 kg ice at -10 degree celsius is mixed with 1 kg water at hundred degree Celsius then find equilibrium temperature and mixture content​

Answer 1

Answer:

The equilibrium mixture is at 9.98 degree C and it contains 2 kg of water

Explanation:

As we know that when two system is mixed then at thermal equilibrium heat given by hot body must be equal to the heat absorbed by cold body

So here heat absorbed by ice = heat given by hot water

let the equilibrium temperature is T degree C

so heat given by water

$Q_1 = ms(100 - T)$

$Q_1 = 1(4186)(100 - T)$

now heat absorbed by ice

$Q_2 = mL + m(4186)(T - 0)$

$Q_2 = 1(335000) + 1(4186)(T - 0)$

So we will have

$Q_1 = Q_2$

$4186(100 - T) = 335000 + 4186 T$

$83600 = 8372 T$

$T = 9.98 ^oC$

Since this temperature is more than 0 degree C so here all ice will convert into water and hence the mixture will contain only (1 + 1) = 2 kg water

#Learn

Topic : Calorimetry

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