Question

Illustration 1.15 A lump of ice of 0.1 kg at -10°C is
put in 0.15 kg of water at 20°C. How much water and ice
will be found in the mixture when it has reached thermal
equilibrium? Specific heat of ice = 0.5 kcal/kg/K and its
latent heat of melting = 80 kcal/kg.​

Answer 1

Explanation:

(HEAT NEEDED TO CONVERT ICE FROM –10°C TO 0°C) +( HEAT NEEDED TO CONVERT INTO WATER AT 0°C)= ms(t2–t1) + mL

= 0.1 * 0.5 * 10 + 0.1 *80

= 8.5 kcal

WE ALSO HAVE WATER AT 20°C ....

HEAT LOST IN CONVERTING WATER FROM 20°C TO 0°C

= ms(t2–t1)

= 0.15 * 1 * 20 = 3 kcal

THIS HEAT IS NOT SUFFICIENT TO CONVERT ALL ICE INTO WATER ......

HENCE ONLY SOME AMOUNT OF ICE WILL MELT INTO WATER....

LET THAT AMOUNT OF ICE MELTED BE x kg

Heat gained by ice = Heat lost by water

x * 0.5 *10 + x*80 = 3

x(5+80) = 3

x = 3/85

x = 0.035 kg

TOTAL WATER AT EQUILIBRIUM = 0.15 + 0.035 kg

= 0.185 kg

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