A professor orders 250 mL of house coffee at precisely 95°C. She then adds enough milk at 10°C to drop the temperature of the coffee to exactly 90° C.
Calculate the amount of milk (in mL) the professor must add to reach this temperature. Assume that the coffee and milk have the same specific heat capacity: 4.184 J/(g × °C). Assume that they also have the same density: 1.0 g/mL.
19.3
mass in grams = mL milk
mass in grams = mL coffee
heat lost by coffee + heat gained by milk = 0
[mass coffee x specific heat coffee x (Tfinal-Tinitial)] + [mass milk x specific heat milk x (Tfinal-Tinitial)] = 0
Substitute the numbers and solve for mass milk in grams which = mL milk.
34
41.8
To solve this problem, we need to use the principle of heat transfer, specifically the equation for heat transfer (Q):
Q = mcΔT
Where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this problem, we want to find the mass (m) of milk that needs to be added. The density of the milk is given as 1.0 g/mL, so the volume (V) of milk can be used in place of mass:
V = m
We are also given that the specific heat capacity (c) of the coffee and milk is 4.184 J/(g × °C).
First, let's calculate the heat transferred when the temperature of the coffee is changed from 95°C to 90°C:
Q1 = mcΔT
= (250 mL) × (4.184 J/g°C) × (90°C - 95°C)
= (250 mL) × (4.184 J/g°C) × (-5°C)
= -523 J
Next, let's calculate the heat transferred when the milk is added to the coffee to bring its temperature down to 90°C:
Q2 = mcΔT
= (1.0 g/mL × V mL) × (4.184 J/g°C) × (10°C - 90°C)
= (1.0 V g) × (4.184 J/g°C) × (-80°C)
= -335.2 V J
Since the heat transferred from the coffee to the milk must be equal to the heat transferred from the coffee when its temperature changes, we can set up the equation:
Q1 = Q2
-523 J = -335.2 V J
To solve for V, we divide both sides of the equation by -335.2 J:
V = -523 J / -335.2 J
= 1.56 mL
Therefore, the professor must add 1.56 mL of milk to the coffee in order to reach a temperature of 90°C.