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.)
The sum of heats added is zero.
n*masscube*Lf+n*masscube*Cwater*(19-0)+masswater*Cw(19-25)=0
mass water= densitywater*154liters=154kg
solve for n.
I've been trying that and I'm still not getting the right answer
To solve this problem, we need to calculate the amount of heat transferred from the water to the ice cubes and then use that information to determine the number of ice cubes needed to reach a temperature of 19°C.
First, let's calculate the initial heat content of the water. The formula for heat content is given by:
Q = mcΔT
Where:
Q = heat content (in calories)
m = mass (in grams)
c = specific heat capacity (in cal/g °C)
ΔT = change in temperature (in °C)
Given:
Mass of water (m) = 154 liters = 154,000 grams
Initial temperature of water (T1) = 25°C
Specific heat capacity of water (c) = 1.0 cal/g °C
Change in temperature (ΔT) = T2 - T1 = 19°C - 25°C = -6°C (since the temperature decreased)
Calculating the initial heat content (Q1) of the water:
Q1 = mcΔT
= (154,000 g) * (1.0 cal/g °C) * (-6°C)
= -924,000 cal
Next, let's calculate the amount of heat required to convert the ice cubes into water at 0°C. This involves calculating the heat of fusion. The formula for heat of fusion is given by:
Q_fusion = mL
Where:
Q_fusion = heat of fusion (in calories)
m = mass (in grams)
L = latent heat of fusion (in cal/g)
Given:
Mass of one ice cube (m) = 28 grams
Latent heat of fusion (L) = 80 cal/g (the heat required to convert 1 gram of ice at 0°C to water at 0°C)
Calculating the heat of fusion (Q_fusion) for one ice cube:
Q_fusion = mL
= (28 g) * (80 cal/g)
= 2240 cal
Now, let's determine the number of ice cubes needed to reach a temperature of 19°C by equating the total heat lost by the water to the heat gained by melting the ice cubes.
Total heat lost by the water = Total heat gained by the ice cubes
-Q1 = n * Q_fusion
Where:
n = number of ice cubes added
Substituting the known values:
924,000 cal = n * 2240 cal
Solving for n:
n = 924,000 cal / 2240 cal
= 412.5
Since we cannot have a fraction of an ice cube, we will round this number to the nearest whole number. Therefore, the physics grad student added approximately 413 ice cubes to the pool to reach a temperature of 19°C.