In order to lower the temperature of 350 mL of water from 24°C to 6°C, we need to calculate the amount of heat energy that needs to be removed from the water. We can use the formula:
Q = mcΔT
where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
The specific heat capacity of water is 4.184 J/g°C. Since the density of water is 1.0 g/mL, the mass of 350 mL of water is 350 g.
Q = (350 g) (4.184 J/g°C) (18°C)
Q = 30,276 J
To remove this much heat energy, we can use ice to cool the water. The heat of fusion of water is 333.55 J/g, which is the amount of heat energy required to melt one gram of ice at 0°C.
Q = (m)(333.55 J/g)
m = Q / 333.55 J/g
m = 30,276 J / 333.55 J/g
m = 90.4 g
So, 90.4 grams of ice at 0°C would have to melt to lower the temperature of 350 mL of water from 24°C to 6°C.