You have 300 g of coffee at 55 degrees celcius. Coffee has the same specific heat as water. How much 10 degrees celcius water do you need to add in order to reduce the coffee's temperature to a more bearable 40 degrees celcius?
You have 300 g of coffee at 55 degrees celcius. Coffee has the same specific heat as water. How much 10 degrees celcius water do you need to add in order to reduce the coffee's temperature to a more bearable 49 degrees celcius?
To find out how much 10 degrees Celsius water you need to add, you can use the principle of heat transfer. The formula for heat transfer is:
Q = mcΔT
Where:
Q is the amount of heat transferred
m is the mass of the substance (in this case, coffee or water)
c is the specific heat capacity of the substance (same for coffee and water)
ΔT is the change in temperature
Since coffee and water have the same specific heat capacity, we can use the same value for c in the formula.
First, let's calculate the heat lost by the coffee when it cools down from 55 degrees Celsius to 40 degrees Celsius:
Q1 = mcΔT1
Where:
m = 300 grams (mass of coffee)
ΔT1 = 55°C - 40°C = 15°C (change in temperature)
Now, let's calculate the heat gained by the water when it warms up from 10 degrees Celsius to the final temperature of the mixture, which is 40 degrees Celsius:
Q2 = mcΔT2
Where:
m = mass of water (we need to find this)
ΔT2 = 40°C - 10°C = 30°C (change in temperature)
Since the heat lost by the coffee is equal to the heat gained by the water (assuming no heat loss to the surroundings), we can set up an equation:
Q1 = Q2
mcΔT1 = mcΔT2
Substituting the known values:
300g * 15°C = m * 30°C
Simplifying the equation:
4500 = 30m
Dividing both sides of the equation by 30:
m = 150g
Therefore, you would need to add 150 grams of 10 degrees Celsius water to the coffee to reduce its temperature to 40 degrees Celsius.