Question

1 kg ice at -10°C is mixed with 1kg water at 100°C. then find equilibrium temperature and mixture content​

Answer 1

Answer:

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Answer 2

The equilibrium temperature is 47.4°C .

Given :

Mass of ice , $m_i = 1\ kg$ .

Initial temperature of ice , $T_i=-10^oC$ .

Mass of water , $m_w = 1\ kg$ .

Initial temperature of water , $T_i=-10^oC$ .

We know , by thermodynamics :

Heat gain = Heat lose

Heat gain by ice to increase temperature to 0°C is :

$H_1=mc\Delta T=1 \times 2100\times 10 =21000\ J.$

Heat gain to convert ice to water :

$H_2=mL=1\times 334=334\ J.$

Let , final temperature of system be t .

Therefore , heat gain to reach temperature t is :

$H_3=mc\Delta T=1\times 4200\times (t-0)=2100t$

Now , sum of heat gain is equal to heat loss .

Therefore ,

$H_1+H_2+H_3=mCw(100-t)\\\\21000+334+4200t=4200(100-t)$

Solving above equation we get :

$t=47.4 ^oC$ .

Also ,

Hence , this is the required solution .

Learn More :

Thermodynamics

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