Answer:
B. 3
Step-by-step 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:
B. 3