Question

1kg ice at -10degree Celsius is mixed with the 1kg water then find out equilibrium temperature and mixture contact

Answer 1

your actual question is -> 1 kg ice at -10°C is mixed with 1 kg water at 100°C. Then find equilibrium temperature and mixture content.

ok, let's try to solve it !

here water is in higher temperatures and ice is in lower temperature so, heat flows from water to ice.

if water lost heat until it 0°C

then, Q = ms∆T

= 1kg × 4200 J/°C/kg × 100°C

= 4.2 × 10^5 J

now, ice heat gained by ice , h = ms'.∆T

= 1kg × 2100J/kg/°C × 10°C

= 2.1 × 10^4 J

now, heat required to melt the ice , h' = mLv

= 1kg × 334000 J/Kg

= 3.34 × 10^5 J

here, h + h' = (3.34 + 0.21) × 10^5 = 3.55 × 10^5 J < H = 4.2 × 10^5 J

so, ice is completely melted and then increased their temperature at T.

ms(100 - T) = 3.55 × 10^5 + msT

or, 4.2 × 10^5 - msT = 3.55 × 10^5 + msT

or, 4.2 × 10^5 - 3.55 × 10^5 = 2msT

or, 0.65 × 10^5 = 2 × 1kg × 4200 × T

or, T = 0.65 × 10^5/8400

= 7.73°C

hence, temperature of mixture is 7.73°C and mixture : water ⇔2kg and ice ⇔0kg