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 use the equation for heat transfer:
Q = mcΔT
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)
ΔT is the change in temperature (in °C)
Since coffee has the same specific heat as water, we can use the same value for c, which is 4.18 J/g°C.
Let's break down the problem step by step:
1. Calculate the initial heat of the coffee:
Q1 = mcΔT1
= (300 g) x (4.18 J/g°C) x (49°C - 55°C)
= -125.4 J
Note that the ΔT1 is negative because we are reducing the temperature of the coffee.
2. Calculate the final heat of the coffee:
Q2 = mcΔT2
= (300 g) x (4.18 J/g°C) x (49°C - 55°C)
= -83.6 J
3. Calculate the heat transferred from the coffee to the water during mixing:
Q1 + Q2 = mcΔT
-125.4 J + (-83.6 J) = (300 g + m) x (4.18 J/g°C) x (49°C - 10°C)
At this point, the negative signs cancel each other out, and we can simplify the equation:
-209 J = (300 g + m) x 39.68 J/g
-209 J / 39.68 J/g = 300 g + m
-5.26 g = 300 g + m
m ≈ -305.26 g
Since we can't have a negative mass of water, the negative sign indicates that the coffee will release 305.26 g of water to the atmosphere during cooling.
So, we need to add approximately 305.26 g of water at 10°C to reduce the coffee's temperature to 49°C.