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 solve this problem, we need to calculate the heat lost by the coffee when it cools down from 55 degrees Celsius to 49 degrees Celsius and find the amount of water needed to absorb this heat.
The formula for heat change is:
Q = mcΔT
Where:
Q = heat change (in Joules)
m = mass (in grams)
c = specific heat (in J/g°C)
ΔT = temperature change (in °C)
Since the coffee's specific heat is the same as water's, we can assume c = 4.18 J/g°C for both coffee and water.
Given:
Initial temperature of the coffee, T1 = 55 °C
Final temperature of the coffee, T2 = 49 °C
Mass of the coffee, m1 = 300 g
Temperature change, ΔT = T2 - T1 = 49 °C - 55 °C = -6 °C (negative because it's a temperature decrease)
Now we can calculate the heat change for the coffee:
Q1 = m1 * c * ΔT
Q1 = 300 g * 4.18 J/g°C * (-6 °C)
Q1 = -7494 J
Since energy is conserved, this amount of heat (Q1) will be absorbed by the water that we add to the coffee.
Let's assume we add x grams of water at 10 °C. The final temperature of the coffee and water mixture will be 49 °C.
The heat gained by the water is given by:
Q2 = m2 * c * ΔT
Q2 = x g * 4.18 J/g°C * (49 °C - 10 °C)
Q2 = 196.8 x g J
Now, since energy is conserved:
Q1 = Q2
-7494 J = 196.8 x g J
Solving for x:
x = -7494 J / 196.8 J/g
x ≈ -38.1 g
We can't have negative mass, so we disregard the negative sign:
x ≈ 38.1 g
Therefore, we need to add approximately 38.1 grams of 10 degrees Celsius water to reduce the coffee's temperature to 49 degrees Celsius.