Specific heat problem simple, but im missing something ?

In summary, the grad student added 624.8 ice cubes to his pool to get the temperature to 16 C, using the given values for the volume of water, density of water, specific heat of water, specific heat of ice, and latent heat of fusion of water. The mistake in the initial calculation was using the specific heat of ice instead of the specific heat of water, as the ice cubes would first melt into liquid water before increasing in temperature.
  • #1
jenzao
48
0
specific heat problem...simple, but I am missing something ?

Homework Statement


Trying to beat the heat of the last summer, a physics grad student went to the local toy store and purchased a plastic child's 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 30g, into the pool. (The ice cubes were originally at 0C.) He continued to add ice cubes, until the temperature stabilized to 16C. 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.)



Homework Equations


m=rho*V
Q=mc delta T
QL = m Lf


The Attempt at a Solution



volume of water: 200L = .2m^3
m = rho*V
m = 1000 * .2 = 200kg of water

Q of water = mc delta T
= .2 * 4.186 * 9
= 7534.8J

Q of ice = mc delta T
= .03kg * 2.09 * 16
= 1.00J

latent heat of fusion for each cube of ice is..
QL = m*Lf
= .03 * (80*4.186)
= 10.05J

so Qwater / (Qice + QL) = number of ice cubes.

7534.8J / (1.00J + 10.05J)

= 681.88 cubes of ice

= wrong..

??please help

(i did notice that this answer is intuitively wrong, but can't see the prob??)
 
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  • #2


That was my first answer as well, but the answer lies in trying to use everything that you are given. Consider Q of ice and the heat of fusion. What is the heat of fusion? What does that heat go into doing? See if that helps you.
 
  • #3


Thanks for the response.

Q of ice is the heat that flows into each icecube to raise its temp from 0 to 16deg
heat of fusion is the heat required to melt an ice cube of mass 30g.
The heat is used first to melt the ice cube (no temp difference)
Then it is used to raise the temp to 16deg
Im using all the info that's given, I still don't get it?
 
  • #4


someone please help. I've done all the work, just missing some small thing and don't know what the prob is?
 
  • #5


The ice cubes are initially at 0C. For the 16C rise, what phase (ice or liquid?) is appropriate? This determines the value of c.
 
  • #6


for the 16deg rise, the phase that is appropriate is water (im not sure i understad the question?)
at zero deg it is ice, so i use the latent heat of fusion value that i am given (80cal/g)
because the ice has to first melt.
then i multiply this by the ice cube mass (.03kg) and 4.186 (to convert to joules)
= 10.05J

then the just melted ice (now water) rises in temp from 0 to 16.
this is all above.
i still don't see the error?
 
  • #7


Okay. You originally used 2.09, the specific heat of ice, instead of 4.186.
 
  • #8


can you tell me exactly where I am wrong, i don't understand.

Q of water = mc delta T
= .2 * 4.186 * 9
= 7534.8J

Q of ice = mc delta T
= .03kg * 2.09 * 16 ( i am using 2.09 because specific heat of ice is 0.5cal/g. Therefore .5*4.186 = 2.09)
= 1.00J

latent heat of fusion for each cube of ice is..
QL = m*Lf
= .03kg * (80*4.186)
= 10.05J

so Qwater / (Qice + QL) = number of ice cubes.

7534.8J / (1.00J + 10.05J)
= 681.88 cubes of ice (this is wrong)
 
  • #9


jenzao said:
can you tell me exactly where I am wrong, i don't understand.

Q of water = mc delta T
= .2 * 4.186 * 9
= 7534.8J

Q of ice = mc delta T
= .03kg * 2.09 * 16 ( i am using 2.09 because specific heat of ice is 0.5cal/g. Therefore .5*4.186 = 2.09)
= 1.00J
This is where the mistake is. The ice melts first into liquid water, and then increases by 16 C in the liquid phase.

Use c = 4.18 J/g/C for liquid water, not 2.09.

Here you were implying that solid ice changes temperature from 0C to 16C.

latent heat of fusion for each cube of ice is..
QL = m*Lf
= .03kg * (80*4.186)
= 10.05J

so Qwater / (Qice + QL) = number of ice cubes.

7534.8J / (1.00J + 10.05J)
= 681.88 cubes of ice (this is wrong)
 
  • #10


oooohhh. that was stupid. yes, thanks for the conceptual explanation. that totally makes sense.
answer = 624.8 cubes of ice.
 

Related to Specific heat problem simple, but im missing something ?

1. What is specific heat and why is it important?

Specific heat is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. It is important because it helps us understand how different substances react to changes in temperature and how much energy is needed to heat or cool them.

2. How is specific heat measured?

Specific heat is usually measured using a calorimeter, which is a device that measures the heat absorbed or released by a substance during a chemical or physical change. The specific heat of a substance can also be calculated using its mass, temperature change, and heat capacity.

3. How does specific heat differ between substances?

Different substances have different specific heats because of their molecular structure and composition. For example, substances with a more complex molecular structure tend to have higher specific heats, meaning they require more energy to change their temperature.

4. What are some real-life applications of specific heat?

Specific heat has many practical applications, such as in cooking and food preservation, climate control in buildings, and in industrial processes such as metal casting and glassmaking. It is also important in understanding and predicting weather patterns and the behavior of different substances in the environment.

5. How can I solve a specific heat problem?

To solve a specific heat problem, you need to know the mass of the substance, its initial temperature, the final temperature, and the specific heat of the substance. From there, you can use the formula Q = mcΔT, where Q is the heat absorbed or released, m is the mass, c is the specific heat, and ΔT is the temperature change. Plug in the known values and solve for the unknown variable.

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