A local barista serves coffee at 85 C. You add ice to the coffee to cool it to 55 C. Assume that an ice cube is 24g and -18.5 degrees Celsius. Hiw many ice cubes would you need to add to your 355mL cup of coffee to bring it to 55 degrees Celsius?.. The specific heat of ice is 2.95J/g degrees Celsius, the specific heat is 4.184 J/g degrees Celsius, and the specific heat of fusion of water is 334 J/g. Remember that an ice cube will need to be warmed to 0 degrees Celsius, will melt, and then the newly melted water will be warmed to 55 degrees Celsius.

B. 3

Explanation:

To decrease the temperature of your coffee from 85°C to 55°C your system need to absorb energy. This energy will be absorbed from the addition of some ice.

How many energy must be absorbed? You can use:

Q = C×m×ΔT

Where Q is heat (Energy), C is specific heat of your solution (4.184J/g°C), m is its mass (mass of 355mL of coffee = 355g) and ΔT is change in temperature (85°C-55°C = 30°C)

Replacing, your ice needs to absorb:

Q = C×m×ΔT

Q = 4.184J/g°C×355g×30°C

Q = 44559.6J

The energy that is taken from an ice cube to change its temperature from -15°C to 55°C is:

Energy from -15°C to 0°C (C of ice = 2.95J/g°C):

Q = C×m×ΔT

Q = 2.95J/g°C×24g×15°C

Q = 1062J

Now the energy taken to pass the ice from solid to liquid is:

Q = ΔHf×m

Q = 334J/g×24g

Q = 8016J

And the energy to increase the temperature of 0°C to 55°C of 24g of ice:

Q = 4.184J/g°C×24g×55°C

Q = 5522.9J

And the total energy that 1 ice cube needs is:

Q = 1062J + 8016J + 5522.9J

Q = 14600.9J

But you need 44559.6J to decrease the temperature of your coffee, that is:

44600J / 14600.9J = 3.05

≈ 3 ice cubes to decrease the temperature of the coffee.

Right solution:

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

c. 75.7

i think that is the right