How much milk at 10.00°C needs to be added to a small 150.0 ml cup of coffee at 95.00 °C to drop the temperature to 90.00°C?
Assume coffee and milk have the same specific heat capacity as water: 4.186 J/g °C or 1 cal/ g °C, and they also have the same density: 1.0 g/mL
[mass milk x specific heat milk x (Tfinal-Tinitial)] + [mass coffee x specific heat coffee x (Tfinal-Tinitial)] = 0
Substitute and solve for mass milk.
To solve this problem, we need to calculate the amount of milk required to drop the temperature of the coffee from 95.00°C to 90.00°C.
The specific heat capacity (C) formula can be used to calculate the heat exchanged between two substances:
Q = m * C * ΔT
Where:
Q is the heat exchanged (in joules or calories)
m is the mass of the substance (in grams or milliliters)
C is the specific heat capacity of the substance (in J/g °C or cal/g °C)
ΔT is the change in temperature (in °C)
Step 1: Calculate the heat exchanged by the coffee using the formula Q = m * C * ΔT.
Q (coffee) = 150.0 ml * 1.0 g/mL * 4.186 J/g °C * (95.00°C - 90.00°C)
Step 2: Calculate the heat exchanged by the milk using the formula Q = m * C * ΔT.
Q (milk) = m (unknown) * 1.0 g/mL * 4.186 J/g °C * (initial temperature - final temperature)
Step 3: Set the heat exchanged by the coffee equal to the heat exchanged by the milk:
Q (coffee) = Q (milk)
150.0 ml * 1.0 g/mL * 4.186 J/g °C * (95.00°C - 90.00°C) = m * 1.0 g/mL * 4.186 J/g °C * (initial temperature - 90.00°C)
Step 4: Solve for the unknown mass (m):
m = (150.0 ml * 1.0 g/mL * 4.186 J/g °C * (95.00°C - 90.00°C)) / (1.0 g/mL * 4.186 J/g °C * (initial temperature - 90.00°C))
Step 5: Calculate the value of m to find the mass of milk needed to drop the temperature.
m = (150.0 ml * (95.00°C - 90.00°C)) / (initial temperature - 90.00°C)
Now you can plug in the value of the initial temperature (10.00°C) to find the mass of milk needed to drop the heat to 90.00°C.