How many grams of coffee must evaporate from 350 g of coffee in a
100-g glass cup to cool the coffee from 95.0ºC to 45.0ºC ? You may
assume the coffee has the same thermal properties as water and that the
average heat of vaporization is 2340 kJ/kg (560 cal/g). (You may neglect
the change in mass of the coffee as it cools, which will give you an
answer that is slightly larger than correct.)
To calculate the grams of coffee that must evaporate, we need to use the formula:
Q = mcΔT
Where:
Q = heat energy
m = mass
c = specific heat capacity
ΔT = change in temperature
In this case, we need to consider the heat required to cool the coffee, which is given by:
Q = mcΔT
The heat required to cool the coffee can be calculated by:
Q = (mass of coffee - mass of evaporated coffee) * specific heat capacity * ΔT
Let's go through the steps:
Step 1: Calculate the heat required to cool the coffee.
Q = 350 g * 4.18 J/g°C * (95 - 45)°C
Q = 350 g * 4.18 J/g°C * 50°C
Q = 73570 J
Step 2: Convert the heat energy to kilojoules (kJ).
1 kJ = 1000 J
73570 J = 73.57 kJ
Step 3: Calculate the mass of the coffee that needs to evaporate.
q = m * hv
Where:
q = heat of vaporization
m = mass
hv = specific latent heat of vaporization
Rearranging the formula, we have:
m = q / hv
Let's substitute the values:
m = 73.57 kJ / 2340 kJ/kg
m ≈ 0.032 g
Therefore, approximately 0.032 grams of coffee must evaporate from the 350 g of coffee in the 100-g glass cup to cool the coffee from 95.0ºC to 45.0ºC.
To find the amount of coffee that must evaporate, we need to consider the heat transfer involved in cooling the coffee.
The formula for heat transfer is as follows:
Q = mcΔT
Where:
Q is the heat transfer (in joules)
m is the mass of the substance (in kg)
c is the specific heat capacity (in J/kg·K)
ΔT is the change in temperature (in K)
In this case, the specific heat capacity of coffee is assumed to be the same as water, which is approximately 4186 J/kg·K.
First, let's convert the given temperature values from Celsius to Kelvin:
Initial temperature, T1 = 95.0ºC + 273.15ºC = 368.15 K
Final temperature, T2 = 45.0ºC + 273.15ºC = 318.15 K
Now let's substitute the values into the formula and rearrange it to solve for mass (m):
Q = mcΔT
Since we know that Q = heat of vaporization = 2340 kJ/kg = 2340 × 10^3 J/kg,
2340 × 10^3 = m × 4186 × (318.15 - 368.15)
Now let's solve for m:
m = (2340 × 10^3) / (4186 × (318.15 - 368.15))
m ≈ 0.071 kg
To convert the mass from kg to grams, multiply by 1000:
m ≈ 71 grams
Therefore, approximately 71 grams of coffee must evaporate from the 350 grams of coffee in the glass cup to cool it from 95.0ºC to 45.0ºC.
the sum of the heats gained is zero.
350cwater*(45-95)+100cglass(45-95)+M*2340=0
look up the cwater, and cglass, an solve for M