Your 300mL cup of coffee is too hot to drink when served at 87C. What is the mass of an ice cube, taken from a -20C freezer, that will cool your coffee to a pleasant 58C?
You want to remove M C (delta T) = 300 g*1.00 Cal/(g C) * 29 C = 8700 calories from the coffee. If m is the mass of ice you add, the heat absorbed by the ice while melting and incresing in temperature from -20 to +58 C is
m*Cice*(20) + mCwater*58 + m*80 Cal/g
The specific heat of ice, Cice = 0.51 Cal/g C
8700 cal = 10.2 m + 58 m + 80 m = 148.2
Solve for m
q1 = heat to move ice from -20 to zero.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial). [Tf will be 0 and Ti will be -20 but watch that sign since it will change.
q2 = heat to melt the ice.
q2 = mass ice x heat fusion.
q3 = heat to move ice from zero to 58.
q3 = mass ice x specific heat water (its now a liquid) x (58-0).
q4 = loss of heat to move 300 mL water from 87 to 58.
q4 = mass water x specific heat water x (58-87).
q1 + q2 + q3 + q4 = 0
You have only one unknown in all of the above which is mass ice. Solve for that. Check my thinking. I won't be on for the next 6 hours or so.
To calculate the mass of the ice cube needed to cool the coffee, we need to utilize the equation for heat transfer:
Q = m * c * ΔT
Where:
Q = heat transfer (in Joules)
m = mass (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
First, let's calculate the amount of heat transferred to the coffee Q1 to cool it from 87°C to 58°C:
Q1 = m1 * c1 * ΔT1
And the amount of heat transferred from the ice cube Q2 to cool it from -20°C to 58°C:
Q2 = m2 * c2 * ΔT2
Since heat lost by the coffee equals heat gained by the ice cube, we can equate the two equations:
Q1 = Q2
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Assuming the specific heat capacity for coffee (c1) is approximately 4.18 J/g°C, and for ice (c2) is approximately 2.09 J/g°C, we can substitute the given values:
m1 * 4.18 * (87 - 58) = m2 * 2.09 * (58 - (-20))
Now we can solve for the mass of the ice cube (m2):
(87 - 58) * 4.18 * m1 = (58 - (-20)) * 2.09 * m2
29 * 4.18 * m1 = 78 * 2.09 * m2
1203.78 * m1 = 162.42 * m2
m2 = (1203.78 * m1) / 162.42
Since the initial volume of the coffee is 300 mL, we can assume the final volume after adding the ice cube is still 300 mL. Assuming the density of water is approximately 1 g/mL, we can substitute the values:
m2 = (1203.78 * 300) / 162.42
m2 = 2219.26 g
Therefore, the mass of the ice cube needed to cool the coffee to 58°C is approximately 2219.26 grams.
To find the mass of the ice cube needed to cool the coffee, we can use the principle of heat transfer. The heat lost by the coffee when it cools down will be equal to the heat gained by the ice cube when it melts.
First, we need to find the heat lost by the coffee to cool from 87°C to 58°C. We can use the heat capacity formula:
Q = m * c * ΔT
where:
Q is the heat lost (or gained),
m is the mass of the substance (coffee),
c is the specific heat capacity, and
ΔT is the change in temperature.
For water (coffee), the specific heat capacity is approximately 4.18 J/g°C.
Using ΔT = 87°C - 58°C = 29°C and the formula, we get:
Q = m_coffee * c_water * ΔT
Next, we need to find the heat gained by the ice cube when it melts. The heat gained (or lost) during a phase change is given by:
Q = m * ΔHfus
where:
Q is the heat gained (or lost),
m is the mass of the substance (ice),
ΔHfus is the enthalpy of fusion (heat of fusion) for water, which is 334 J/g.
Since the ice cube will be initially at -20°C and needs to melt to 0°C, the change in temperature (ΔT) is 20°C.
Q = m_ice * ΔHfus
Now, we equate the heat lost by the coffee to the heat gained by the ice cube:
m_coffee * c_water * ΔT = m_ice * ΔHfus
We can rearrange the equation to solve for m_ice:
m_ice = (m_coffee * c_water * ΔT) / ΔHfus
Now we can calculate the mass of the ice cube. Given that the volume of the coffee is 300 mL, we can assume that its mass is approximately 300 g (since the density of water is approximately 1 g/mL). Substituting the values into the equation:
m_ice = (300 g * 4.18 J/g°C * 29°C) / 334 J/g
m_ice ≈ 92.1 g
Therefore, you would need an ice cube with a mass of approximately 92.1 grams to cool your coffee from 87°C to 58°C.