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 answer this question, we need to calculate the heat transfer that occurs when mixing the hot coffee with cooler water. The equation we will use is:
Q = mcΔT
Where:
Q = Heat transfer (in Joules)
m = Mass of the substance (in kilograms)
c = Specific heat capacity (in J/g°C or J/kg°C)
ΔT = Change in temperature (in °C)
First, we need to convert the mass of the coffee from grams to kilograms:
300 g = 0.3 kg
Now, let's calculate the heat transfer when cooling down the coffee from 55°C to 49°C. Since coffee has the same specific heat as water, we'll use the specific heat capacity of water, which is approximately 4.18 J/g°C:
Q = (0.3 kg) * (4.18 J/g°C) * (55°C - 49°C)
Q = 0.3 kg * 4.18 J/g°C * 6°C
Q = 7.53 Joules
Now, let's calculate the heat transfer when warming up the water from 10°C to the final temperature when mixed with the coffee (ΔT).
Q = mcΔT
We don't know the mass of the water yet, so let's assume it is represented by "m_w" (in kilograms). Since water has the same specific heat as coffee, we'll use the specific heat capacity of water again:
Q = (m_w kg) * (4.18 J/g°C) * (ΔT °C)
Given that we want the final temperature to be 49°C, the change in temperature (ΔT) is:
ΔT = 49°C - 10°C
ΔT = 39°C
Now, substituting back into the equation:
7.53 J = (m_w kg) * (4.18 J/g°C) * (39°C)
By canceling out the units, we find:
7.53 = m_w kg * 4.18 * 39
Simplifying the equation, we can solve for the mass of water needed:
m_w kg = 7.53 / (4.18 * 39)
Calculating this expression, we find:
m_w kg ≈ 0.046 kg
Therefore, you would need approximately 0.046 kg (or 46 grams) of water at 10°C to mix with the 300 grams of coffee at 55°C in order to reduce the temperature to a more bearable 49°C.