How much ice in grams would have to melt to lower the temperature of 58.2 g of water from 55.0 degrees C to 0 degrees C? Assume the density of water is 1.0 g/mL

Any advice on how to tackle this problem would be appreciated.

Thanks

How much heat must be removed? That is

q = mass H2O x specific heat H2O x (Tfinal-Tinitial)

How much can 1 g ice remove? That's the heat fusion.
g ice = q/heat fusion
About 40 grams?

To tackle this problem, we need to find the amount of heat that needs to be lost from the water in order to lower its temperature from 55.0°C to 0°C. This heat loss can be calculated using the equation:

Q = m × c × ΔT

Where:
Q is the heat lost (in calories or joules),
m is the mass of the water (in grams),
c is the specific heat capacity of water (in calories per gram-degree Celsius or joules per gram-degree Celsius),
ΔT is the change in temperature (in degrees Celsius).

Since we are given the mass of the water (58.2 g) and the change in temperature (55.0°C to 0°C), we need to determine the specific heat capacity of water.

The specific heat capacity of water is approximately 1.0 cal/g°C or 4.18 J/g°C.

Let's substitute the values into the equation:

Q = 58.2 g × 1.0 cal/g°C × (0 - 55.0°C)
Q = -3191.0 cal

To convert calories to grams of ice, we need to use the heat of fusion of ice, which is the amount of heat required to change 1 gram of ice into water at 0°C. The heat of fusion of ice is approximately 79.7 cal/g.

Let's substitute the values into the equation:

Q = m × ΔH_fusion

-3191.0 cal = m × 79.7 cal/g

Rearranging the equation to solve for m:

m = -3191.0 cal / 79.7 cal/g
m ≈ 40.0 g

Therefore, approximately 40.0 grams of ice would have to melt to lower the temperature of 58.2 g of water from 55.0°C to 0°C.