lemons. When finished, there are 221 grams of lemonade at 24.8°C. Chris doesn\'t

Question

lemons. When finished, there are 221 grams of lemonade at 24.8°C. Chris doesn't like ice in lemonade! Therefore, just enough ice is used to cool the lemonade to 10.1°C. Of course, the ice will melt and reach the same temperature. If the ice starts at -10.6 C, and if the specific heat of lemonade is the same as that of water, Chris use? Assume there is no heat transfer to or from the surroundings. hat is not a very refreshing temperature, so it must be cooled with ice. But It is a hot summer day and Chris wants a glass of lemonade. There is none in the refrigerator, so a new batch is prepared from freshly squeezed Data for water at 1 atm: Melting point-0.00C Specific heat liquid -4.18 J/g. C C Specific heat solid -2.06 J/g Heat of fusion- 333 J/g

Explanation / Answer

Ans. Let the required mass of ice is X gram.

#Step 1: Heat absorbed by solid ice to reach 0.00C from -10.60C

The required Heat gain is given by-

q = m s dT                            - equation 1

Where,

q = heat

m = mass

s = specific heat

dT = Final temperature – Initial temperature

Putting the values in equation 1-

            q1 = X g x (2.06 J g-10C-1) x [0.00C – (-10.60C)]

            Hence, q1 = 21.836X J

# Step 2: Heat absorbed by solid ice at 0.00C to from liquid water at 0.00C

Heat absorbed by ice at 0.00C to melt at same temperature is given by-

            q2 = m x C                - equation 2

            Where, C = heat of fusion

            Or, q2 = X g x (333.0 J/ g)

            Hence, q2 = 333X J

#Step 3: Heat absorbed by liquid water at 0.00C to reach 10.10C

Putting the values in equation 1-

            q3 = X g x (4.184 J g-10C-1) x (10.10C – 0.00C)

            Hence, q3 = 42.2584X J

# Step 4: Heat lost by lemonade to cool to 10.10C

Putting the values in equation 1-

            q4 = 221.0 g x (4.184 J g-10C-1) x (10.10C – 24.80C)

            Hence, q4 = - 13592.5608 J

# Step 5: Total heat gained by ice to reach thermal equilibrium with lemonade must be equal to the total heat lost by the lemonade.

So,

            -q4 = q1 + q2 + q3

            Or, - (- 13592.5608 J) = 21.836X J + 333X J + 42.2584X J

            Or, 13592.5608 J = 397.0944 X J

            Or, X = 13592.5608 / 397.0944

            Hence, X = 34.23

Therefore, required mass of ice = X grams = 34.23 g

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