Assume the optimal coffee drinking temperature is 75°C, and further assume coffee has the same heat capacity and density as water.

a. How much energy is required to raise the temperature of 1 cup (0.237 liters) of coffee initially at 25°C to the optimal temperature?

b. How much energy is required to heat the coffee to 95°C?

c. The coffee is cooled back down so that the mixture of coffee and melted ice has the optimal temperature. If the ice started at ‐3°C, what mass (m=?) of ice is required?

d. If the water used to make the ice started at 25°C, how much energy was required to freeze it and cool the ice to ‐3°C in the first place?

e. How much energy was wasted by overheating the coffee? This will be the overheating energy, plus the energy to make the ice, less the energy to heat it to the optimal temperature. What is the ratio of wasted energy to useful energy?

I will be happy to critique your thinking.

you have posted these questions under several names, I can only assume you are answer grazing.

My thinking:

a. Q=(4.18)(237)(50)=49533 J

b. Q=(4.18)(237)(70)=69346.2 J

c. Q=(4.18)(237)(Tf-95)+(4.18)m(3) + m(334) +cm(Tf)=0

d. Q=(4.18)(237)(-25) + (237)(334) + (4.18)(237)(-3)

e. Don't know

a. To calculate the energy required to raise the temperature of the coffee to the optimal temperature of 75°C, you can use the equation:

Energy = mass * specific heat capacity * temperature difference

First, let's convert the volume of coffee from liters to grams. Assuming the density of coffee is the same as water (1 gram per milliliter):

Volume of coffee = 0.237 liters = 237 milliliters = 237 grams

Next, we need to calculate the temperature difference:
Temperature difference = 75°C - 25°C = 50°C

The specific heat capacity of water is approximately 4.18 J/g°C.

Plugging in the values into the equation:

Energy = 237 grams * 4.18 J/g°C * 50°C
Energy = 495,210 J

Therefore, approximately 495,210 Joules of energy is required to raise the temperature of 1 cup (0.237 liters) of coffee from 25°C to 75°C.

b. To calculate the energy required to heat the coffee to 95°C, we use the same equation as in part a, but with a different temperature difference:

Temperature difference = 95°C - 25°C = 70°C

Energy = 237 grams * 4.18 J/g°C * 70°C
Energy = 694,884 J

Therefore, approximately 694,884 Joules of energy is required to raise the temperature of 1 cup (0.237 liters) of coffee from 25°C to 95°C.

c. To find the mass of ice required to cool the coffee down to its optimal temperature of 75°C, we need to use the equation:

Energy gained by ice = Energy lost by coffee

Since the ice starts at -3°C and ends at 75°C, the temperature difference is:

Temperature difference = 75°C - (-3°C) = 78°C

The specific heat capacity of ice is approximately 2.09 J/g°C.

Let's assume the ice melts completely and absorbs all the heat energy released by the coffee. The energy gained by the ice can be calculated as:

Energy gained by ice = mass of ice * specific heat capacity of ice * temperature difference

Assuming the coffee loses the same amount of energy as in part a:

Energy lost by coffee = 495,210 J

Equating the two equations:

mass of ice * 2.09 J/g°C * 78°C = 495,210 J

Simplifying:

mass of ice = 495,210 J / (2.09 J/g°C * 78°C)

mass of ice ≈ 3,210 grams

Therefore, approximately 3,210 grams (or 3.21 kilograms) of ice is required to cool the coffee down to its optimal temperature.

d. To calculate the energy required to freeze the water and cool it to -3°C, we need to consider two steps. First, we calculate the energy required to freeze the water from 25°C to 0°C. Then, we calculate the energy required to cool the frozen water from 0°C to -3°C.

The specific heat capacity of water is still 4.18 J/g°C.

To freeze the water from 25°C to 0°C, we use the equation:

Energy = mass of water * specific heat capacity of water * temperature difference

For this step, the temperature difference is:

Temperature difference = 0°C - 25°C = -25°C (negative because we are cooling the water)

Next, we need to calculate the energy required to cool the frozen water from 0°C to -3°C:

Energy = mass of frozen water * specific heat capacity of ice * temperature difference

For this step, the temperature difference is:

Temperature difference = -3°C - 0°C = -3°C

Adding the energies from both steps will give us the total energy required:

Total energy = Energy required to freeze water + Energy required to cool frozen water

Use the equation similar to parts a and c to calculate these energies.

e. To calculate the ratio of wasted energy to useful energy, we need to consider the total energy wasted (overheating energy plus energy to make the ice) and the useful energy (energy required to heat the coffee to optimal temperature).

The energy wasted can be calculated as follows:

Energy wasted = Energy used to make ice + Energy overheating coffee

Energy used to make ice = mass of ice * specific heat capacity of ice * temperature difference (from part c)
Energy overheating coffee = Energy required to heat coffee to 95°C - Energy required to heat coffee to 75°C (from parts a and b)

The useful energy can be calculated as:

Useful energy = Energy required to heat coffee to 75°C

Finally, calculate the ratio of wasted energy to useful energy:

Ratio = Energy wasted / Useful energy