Question

I recall on more than one occasion when our kids were small, we would come inside from playing in the snow and I would make hot chocolate by heating the milk in the microwave. Invariably, I would get the milk too hot for at least one of them. Rather than adding more cold milk to the hot chocolate, I dropped a couple of small ice cubes into the mug. I figured they would never notice the dilution effect that the water had on the hot chocolate since they always enjoyed adding about thirty marshmallows to the mixture after I served it to them anyway.

Although I never actually took any temperatures of the hot chocolate, let’s assume that the initial temperature was 175⁰F and the mug contained 350 g of hot chocolate.

a)   If the initial temperature of the hot chocolate was is 175⁰F, what would the temperature of the drink be once the ice cubes had completely melted and the mixture stirred? Express your answer in “degrees Celsius” to one decimal place. Since the marshmallows were added after I served the drinks, they would have no impact on the temperature of the drink at the time I served it. Additional information is provided below.

b)   From a purely personal perspective (i.e., in your own opinion), would you consider the chocolate milk still be too hot for you personally to drink or would it be a bit on the cool side? How can you support your answer?  

Additional information:

Weight of ice cubes   =   75 grams total weight
Initial temperature of ice cubes   =   - 18⁰C
Specific heat capacity of the hot chocolate   =   3.751 kJ/kg C⁰

Hints: While this problem can be a little bit “scary” at first, it should not really be that bad if you break the problem down into a series of more manageable steps. Think about what is happening before you charge ahead and start trying to actually solve this problem.  

You can simplify things greatly if you draw a diagram to represent what is happening. Draw a vertical line to represent a “thermometer” and mark the important points on it (this diagram will also assist you in determining the temperature changes etc.). These points will include the starting temperatures of the chocolate milk and the ice as well as the freezing / melting point of the ice.  

Let the final temperature of the thawed ice and milk be X (the chocolate milk and the melted ice will both have the same final temperature since they are mixed together in the same mug). X will be somewhere between the thawing / freezing point of the ice and the initial temperature of the milk - the trick is to find out what the temperature actually is. Remember that the heat lost by the milk will be equal to the heat gained by the ice, if we assume that no heat is lost to the surroundings (which we will do in this case).

Answer 1

==> (175°F-32)×5/9 = 79.44°C

a)    mug contained 350 g of hot chocolate.

Weight of ice cubes   =   75 grams total weight
Initial temperature of ice cubes   =   - 18⁰C
Specific heat capacity of the hot chocolate   =   3.751 kJ/kg C⁰ or 3.751 J/g C⁰

temperature of the drink be once the ice cubes had completely melted and the mixture stirred :

heat capacity of ice = 2.094J/g C⁰

heat capacity of water = 4.184J/g C⁰

Q(drink) = 350 g * 3.751 J/g C⁰ * ( T - 79.44°C)

Q(ice) = 75 g * 2.094 J/g C⁰ * ( 0 + 18°C) = 2826.9 J

Q(fusion) = 75 g* 333.89 J = 25041.67 J

Q (water) = 75 g * 4.184 J/g C⁰ * ( T - 0 C)

we have ,

- Q(drink) = Q (ice) + Q (fusion) + Q ( water)

350 g * 3.751 J/g C⁰ * ( 79.44°C - T) = 75 g * 4.184 J/g C⁰ * ( T - 0 C) + 27868.57 J

104292.8 J - 1312.85 *T J/C = 313.8 *T J/C + 27868.57 J

1626.65 *T J/C   = 76424.23 J

T = 47.0 C

b) 47 C is OK temperature to drink , hot chocolate as human body temperature 37 C.