PLEASE help because I'm really confused on how to do the entire thing. I don't know how to find the mL. Am I suppose to do anything with the 250mL?
A person stops in every day and orders 250 mL of house coffee at precisely 95°C. but enough milk at 10°C to drop the temperature of the coffee to exactly 90° C.
Calculate the amount of milk (in mL) the we 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.
In these problems one material loses heat and the other gains heat. In this case, the warm coffee loses heat and the cold milk gains heat. All of that exchange must be zero.
heat lost by coffee + heat gained by milk = 0
heat lost by coffee is
mass coffee x specific heat coffee x (Tfinal-Tintial)
heat gained by milk is
mass milk x specific heat milk x (TfinalTinitial)_
You put those together like this
[mass coffee x sp.h. coffee x (Tf-Ti)] + [mass milk x sp.h. milk x (Tf-Ti) = 0
Now you substitute the numbers into the above. We are assuming milk/coffee are really water with sp.h of each = 4.184 J/g*C. And since the density of water is 1.0 g/mL then the 250 mL coffee will be 250 g and the mL milk will be that many grams milk.
[250g coffee x 4.184 J/g*C x (Tf-Ti)] + [mass milk x 4.184 J/g*C x (Tf-Ti)] = 0
Ti for coffee is 95 C
Tf for coffee is 90 C
Ti for milk is 10 C
Tf for milk is 90 C (Tfinal for milk must be the same as Tfinal for coffee since they are mixed.)
The only unknown in that equation is mass milk (in grams) and with the density being 1.0 g/mL that will be the number of mL. If you post your work I'll be happy to check you answer.
To determine the amount of milk (in mL) needed to reach the desired temperature, we can use the concept of heat transfer. The equation for heat transfer is:
Q = m * c * ΔT
Where:
Q is the heat transferred (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/(g × °C)),
and ΔT is the change in temperature (in °C).
In this case, the heat transferred from the coffee to the milk is equal to the heat absorbed by the milk. We need to calculate the mass of the milk.
To find the mass, we can rearrange the equation for density:
density = mass / volume
Rearranging this equation gives us:
mass = density * volume
Since the density of both the coffee and milk is 1.0 g/mL, the mass of the milk is equal to its volume in mL.
To calculate the heat transfer, we'll use the initial temperature of the coffee (95°C) and the final temperature of the mixture (90°C). The specific heat capacity (c) is 4.184 J/(g × °C), as given in the question.
Now let's put it all together to find the amount of milk (in mL) needed:
1. Calculate the heat transferred from the coffee to the milk:
Q = m_coffee * c * (T_final - T_initial)
= 250 mL * 1.0 g/mL * 4.184 J/(g × °C) * (90°C - 95°C)
2. Calculate the heat absorbed by the milk (since Q_coffee = Q_milk):
Q = m_milk * c * (T_final - T_initial)
3. Solve for the mass of the milk:
m_milk = Q / (c * (T_final - T_initial))
Since the density of the milk is 1.0 g/mL, the mass of the milk in grams is equal to the volume of the milk in mL.
So, to find the amount of milk in mL, you can now substitute the values into the equation and solve for m_milk.