A chemistry professor has a cup of coffee containing 50.0mL of room temperature coffee at
25.0C. The professor also has a new pot of hot coffee at temperature of 96.0C. What volume of hot
coffee will the professor need to add to his cold coffee to reach the ideal drinking temperature of
82.0C? Assume that no heat is lost to the coffee cup or to the environment. Also assume that coffee
has the same density (1.0g/mL) and heat capacity (4.184 J/gC) as water.
Use the same formula as in the warm/hot water problem but this time you know Tf and Ti and you want to solve for mass hot coffee.
How do you find Q for this example?
I did
50mL x 4.184 x (82-25) = Heat gained = 11924J
Heat gained 11924J= Heat lost 11924J
Heat lost = 11924J = Mass x 4.184 x (82-96)
11924J / (4.184 x -14) = Mass
Mass = 203.56 = 204 mL (with proper significant figures)
Is this right?
To solve this problem, we need to use the equation for heat transfer:
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 case, we don't know the mass of the hot coffee we need to add, but we do know the volume and density of the coffee. Since density is mass divided by volume, we can rearrange the equation to find the mass:
mass = density * volume
For the hot coffee, we can use the equation:
Q_hot = m_hot * c_hot * ΔT_hot
where Q_hot is the heat transferred by the hot coffee, m_hot is the mass of the hot coffee, c_hot is the specific heat capacity of hot coffee (same as water), and ΔT_hot is the change in temperature for hot coffee.
For the cold coffee, we can use the equation:
Q_cold = m_cold * c_cold * ΔT_cold
where Q_cold is the heat transferred by the cold coffee, m_cold is the mass of the cold coffee, c_cold is the specific heat capacity of cold coffee (same as water), and ΔT_cold is the change in temperature for cold coffee.
Since no heat is lost to the coffee cup or the environment, the heat transferred from the hot coffee to the cold coffee will be equal to the heat transferred from the cold coffee to reach the desired temperature. Therefore, we can set up the equation:
Q_hot = Q_cold
m_hot * c_hot * ΔT_hot = m_cold * c_cold * ΔT_cold
Now we can substitute the known values into the equation:
(96.0 - 82.0) * m_hot * 4.184 = (82.0 - 25.0) * (50.0 + m_hot) * 4.184
Simplifying the equation, we get:
14.0 * m_hot = 57.0 * (50.0 + m_hot)
Next, we can solve for m_hot, the mass of the hot coffee:
14.0 * m_hot = 57.0 * 50.0 + 57.0 * m_hot
14.0 * m_hot - 57.0 * m_hot = 57.0 * 50.0
-43.0 * m_hot = 57.0 * 50.0
m_hot = (57.0 * 50.0) / -43.0
m_hot ≈ -66.51 g
Since mass cannot be negative, this indicates that we do not need to add any hot coffee to the cold coffee. The professor's 50.0 mL of cold coffee at 25.0C will reach the desired temperature of 82.0C without any additional hot coffee.