Trying to beat the heat of summer, a physics grad student went to the local toy store and purchased a child's plastic swimming pool. Upon returning home, he filled it with 154 liters of water at 25°C. Realizing that the water would probably not be cool enough, he threw ice cubes from his refrigerator, each of mass 28 g, into the pool. (The ice cubes were originally at 0°C.) He continued to add ice cubes until the temperature stabilized at 19°C. He then got in the pool.

The density of water is 1000 kg/m3, the specific heat of water is 1.0 cal/g °C, the specific heat of ice is 0.5 cal/g °C, and the latent heat of fusion of water is 80 cal/g.

How many ice cubes did he add to the pool to get the temperature to 19°C? (Consider the pool and ice cubes an isolated system.)
Physics - bobpursley, Saturday, November 20, 2010 at 4:06pm
The sum of heats added is zero.

Help:

n*masscube*Lf+n*masscube*Cwater*(19-0)+masswater*Cw(19-25)=0

mass water= densitywater*154liters=154kg

solve for n.

I've tried to do this from the help above, but I am still not getting the right answer.

check your units, make certain all units are in SI (cal/g) is not SI.

To solve this problem, you need to use the principle of energy conservation. According to the principle, the sum of the heat added to the system should be equal to zero.

Let's break down the equation that needs to be solved step by step:

n * masscube * Lf + n * masscube * Cwater * (19 - 0) + masswater * Cw * (19 - 25) = 0

Here's what each part of the equation represents:

- n * masscube * Lf: This term represents the heat added to the system due to the ice cubes melting. "n" is the number of ice cubes added, "masscube" is the mass of each ice cube (28 g), and "Lf" is the latent heat of fusion of water (80 cal/g). Multiplying these values gives us the total heat added due to the ice cubes melting.

- n * masscube * Cwater * (19 - 0): This term represents the heat gained by the water due to the ice cubes. "Cwater" is the specific heat of water (1.0 cal/g °C), and (19 - 0) is the temperature difference between the ice cubes and the final temperature of the water (19°C). Multiplying these values gives us the heat gained by the water from the ice cubes.

- masswater * Cw * (19 - 25): This term represents the heat lost by the water due to cooling from 25°C to 19°C. "masswater" is the mass of the water in the pool (154 kg), "Cw" is the specific heat of water (1.0 cal/g °C), and (19 - 25) is the temperature difference. Multiplying these values gives us the heat lost by the water.

The equation is set to zero because energy is conserved; the total heat added to the system should be equal to the total heat lost.

To solve for "n," substitute the known values into the equation and then solve for "n":

n * 28 * 80 + n * 28 * 1.0 * 19 + 154 * 1.0 * (19 - 25) = 0

Simplifying the equation:

2240n + 532n + 154(-6) = 0
2772n - 924 = 0
2772n = 924
n = 924 / 2772
n ≈ 0.333

So, the grad student added approximately 0.333 or 1/3 of an ice cube to get the temperature to stabilize at 19°C.