Ice cubes added to pool to make it cool enough

[confused1]

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 200 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 30 g, into the pool. (The ice cubes were originally at 0 C.) He continued to add ice cubes until the temperature stabilized at 16 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 16 C? (Consider the pool and ice cubes an isolated system.)


HELP: Heat lost by water = heat gained by ice cubes. No heat is lost to the surroundings.

HELP: Since the water (subsystem 1) is at a higher temperature, heat will be lost to the ice cubes (subsystem 2). Calculate the heat H that the water gave up from 25 C to 16 C; calculate the heat h that each ice cube gained from 0 C to 16 C including melting. Then the number of ice cubes equals H/h.

 

[Pengwuino]

Using the specific heat of water, calculate how many calories will need to be transferred from the 200 liters of water (convert liters to grams... i think 1 ml is 1 gram of water... not sure though) the 25 - 16 = 9C the guy wants.

Then determine how many calories each ice cube adds by using the latent heat of fusion of water plus the specific heat of water to reach teh 16C. Then figure out how many ice cubes undergoing this process will be required to make up the 9C in teh 200l of water.

 

[thepatientmental]

To determine the number of ice cubes added, we need to first calculate the total heat lost by the water and the total heat gained by the ice cubes. This can be done using the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature.

For the water, the initial temperature is 25 C and the final temperature is 16 C, so ΔT = 9 C. The mass of the water is given as 200 liters, which is equivalent to 200 kg. The specific heat of water is 1.0 cal/g C, so we need to convert the mass to grams. This gives us a total heat lost by the water of Q = (200,000 g)(1.0 cal/g C)(9 C) = 1,800,000 cal.

For the ice cubes, we need to consider both the change in temperature and the melting of the ice. The ice cubes start at 0 C and end at 16 C, so ΔT = 16 C. The mass of each ice cube is given as 30 g. The specific heat of ice is 0.5 cal/g C, so the heat gained by each ice cube is Q = (30 g)(0.5 cal/g C)(16 C) = 240 cal.

However, we also need to consider the heat gained from the melting of the ice cubes. The latent heat of fusion of water is 80 cal/g, so the total heat gained from melting is h = (30 g)(80 cal/g) = 2,400 cal.

Now, we can calculate the number of ice cubes by dividing the total heat lost by the water (1,800,000 cal) by the total heat gained by each ice cube (240 cal + 2,400 cal = 2,640 cal). This gives us a total of 682.7 ice cubes. Since we cannot have a fraction of an ice cube, the grad student likely added 683 ice cubes to the pool in order to lower the temperature to 16 C.