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 principle of heat transfer. The heat lost by the coffee will be equal to the heat gained by the water.
First, let's calculate the heat lost by the coffee. The formula for heat transfer is Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Given:
Initial temperature of coffee (T1) = 55 degrees Celsius
Desired final temperature of coffee (T2) = 49 degrees Celsius
Mass of coffee (m1) = 300 g
Specific heat of coffee (c1) = specific heat of water = 1 calorie/g°C (or 4.18 J/g°C)
Heat lost by the coffee (Q1) = m1 * c1 * (T1 - T2)
Next, let's calculate the heat gained by the water. Since the water is initially at 10 degrees Celsius, it needs to be heated to reach the final temperature of the coffee.
Final temperature of water (T3) = T2 = 49 degrees Celsius
Heat gained by the water (Q2) = m2 * c1 * (T3 - T1)
Since Q1 = Q2 (according to the principle of heat transfer), we can equate the two equations:
m1 * c1 * (T1 - T2) = m2 * c1 * (T3 - T1)
Now, we want to find the mass of water (m2) that needs to be added, so we can rearrange the equation:
m2 = (m1 * (T1 - T2)) / (T3 - T1)
Substituting the given values:
m2 = (300 g * (55°C - 49°C)) / (49°C - 55°C)
Simplifying the equation:
m2 = (300 g * 6°C) / (-6°C)
The -6°C in the denominator arises due to the difference in signs between the temperatures in the equation. It represents the fact that the water needs to be heated from 10°C to reach 49°C, which is an increase in temperature.
m2 = (300 g * 6°C) / (-6°C)
m2 = -1800 g°C / -6°C
m2 = 300 g
Therefore, you need to add 300 grams of 10 degrees Celsius water to reduce the coffee's temperature to 49 degrees Celsius.