Question

You pour a glass of orange juice and realize it is warm (24 Celsius). You find the mass of the orange juice to be 642 grams. A quick search shows the specific heat of orange juice is about 3770J/(kg*C). You add ice cubes from the freezer which you assume are at 0 Celsius. if you add a single 50 gram cube of ice, what is the final temperature of the mixture? How many ice cubes do you need to add before the juice reaches 0 celsius?

Answer 1

Step 1:

Suppose equilibrium temperature is T, then

Heat released by orange Juice = heat gained by ice

Q1 = Q2

m1*C1*dT1 = mi*Lf + mi*Cw*dT2

dT1 = 24 - T

dT2 = T - 0 = T

m1 = mass of orange juice = 642 gm = 0.642 kg

C1 = Specific heat of orange juice = 3770 J/kg-C

mi = mass of ice cubes = 50 gm = 0.05 kg

Lf = Latent heat of fusion of ice = 3.34*10^5 J/kg

Cw = Specific heat capacity of water = 4186 J/kg-C

Using known values:

0.642*3770*(24 - T) = 0.050*3.34*10^5 + 0.050*4186*(T - 0)

Solving above equation:

T = [0.642*3770*24 - 0.050*3.34*10^5]/(0.642*3770 + 0.050*4186)

T = 15.74 C = final temperature of system when one ice cube is added

Part B.

When final temperature of system is 0 C, then

total energy released by orange juice will be:

Q = m1*C1*dT1

Q = 0.642*3770*(24 - 0)

Q = 58088.16 J

Now number of ice cubes required to abosorb above energy will be:

Q1 = energy released by n ice cube during phase change = n*mi*f

Since Q1 = Q, So

n = Q/(mi*f)

n = 58088.16/(0.050*3.34*10^5)

n = 3.478 = 3.5 ice cubes

In integer value:

n = 4 ice cubes required

Let me know if you've any query.