While studying for this quiz you realize that you still have 100 g of lukewarm coffee at 40°C left in a paper cup. When you pour 50 g of boiling water into the cup, the temperature of the resulting coffee-like mixture will now be??
Well, let me do some quick math while juggling some coffee beans... Ah-ha, got it! So, if you take 100 g of lukewarm coffee at 40°C and add 50 g of boiling water, you'll end up with a new drink that's perfect for those who can't decide between lukewarm and boiling. The resulting temperature of the mixture will be "lukewarm-boiling-y" degrees! Though, in all honesty, it would be around... *drum roll*... some degree between 40°C to 100°C, depending on how much each drink cools or heats the other. So, sip it cautiously and be prepared for that unique temperature surprise! Enjoy! 🍵😄
To calculate the final temperature of the resulting coffee-like mixture, we can apply the principle of heat transfer, known as the heat equation:
Heat gained by the lukewarm coffee = Heat lost by the boiling water
The heat gained or lost can be calculated using the specific heat capacity equation:
Q = mcΔT
Where:
Q = heat gained or lost
m = mass of the substance
c = specific heat capacity
ΔT = Change in temperature
For the lukewarm coffee:
m1 = 100 g
c1 = specific heat capacity (varies for different substances, let's assume it to be the same as water, which is 4.18 J/g°C)
ΔT1 = final temperature - initial temperature = final temperature - 40°C
For the boiling water:
m2 = 50 g
c2 = specific heat capacity (same as the lukewarm coffee)
ΔT2 = final temperature - initial temperature = boiling point (100°C) - initial temperature
Now we can set up the equation:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
(100 g) * (4.18 J/g°C) * (final temperature - 40°C) = (50 g) * (4.18 J/g°C) * (100°C - final temperature)
Now we can solve for the final temperature:
(100 g) * (4.18 J/g°C) * final temperature - (100 g) * (4.18 J/g°C) * 40°C = (50 g) * (4.18 J/g°C) * 100°C - (50 g) * (4.18 J/g°C) * final temperature
(100 g) * (4.18 J/g°C) * final temperature + (50 g) * (4.18 J/g°C) * final temperature = (50 g) * (4.18 J/g°C) * 100°C + (100 g) * (4.18 J/g°C) * 40°C
Now we can simplify the equation and solve for the final temperature:
(100 g * 4.18 J/g°C + 50 g * 4.18 J/g°C) * final temperature = (50 g * 4.18 J/g°C * 100°C) + (100 g * 4.18 J/g°C * 40°C)
(418 J/°C + 209 J/°C) * final temperature = (50 g * 4.18 J/g°C * 100°C) + (100 g * 4.18 J/g°C * 40°C)
(627 J/°C) * final temperature = (2090 J + 1672 J)
(627 J/°C) * final temperature = 3762 J
final temperature = 3762 J / 627 J/°C
final temperature ≈ 6°C
Therefore, the temperature of the resulting coffee-like mixture will be approximately 6°C.
To determine the temperature of the resulting coffee-like mixture, we can use the principle of heat transfer. The formula for calculating the final temperature of a mixture is:
Q1 + Q2 = m1 * c1 * ΔT1 + m2 * c2 * ΔT2
Where:
Q1 and Q2 are the amounts of heat transferred from the first and second substances,
m1 and m2 are the masses of the first and second substances,
c1 and c2 are the specific heat capacities of the first and second substances, and
ΔT1 and ΔT2 are the temperature changes of the first and second substances.
In this situation, let's assume the specific heat capacity of coffee is the same as water, which is approximately 4.18 J/g°C. The initial temperature of the coffee is 40°C, and the initial temperature of the boiling water is 100°C.
We can calculate the heat transferred from the coffee as follows:
Q1 = m1 * c1 * ΔT1 = 100 g * 4.18 J/g°C * (40°C - Tf)
We can calculate the heat transferred from the boiling water as follows:
Q2 = m2 * c2 * ΔT2 = 50 g * 4.18 J/g°C * (100°C - Tf)
Since heat is conserved in this process (assuming no heat is lost to the surroundings), Q1 + Q2 must equal zero. We can set up the equation as follows:
100 g * 4.18 J/g°C * (40°C - Tf) + 50 g * 4.18 J/g°C * (100°C - Tf) = 0
Simplifying the equation:
418 * (40 - Tf) + 209 * (100 - Tf) = 0
Solving this equation will give us the final temperature (Tf) of the resulting coffee-like mixture.
heat in = heat out
100 (T-40) = 50 (100-T)
2 T - 80 = 100 - T
3 T = 180