Help with Water, Ice, and metal finding temperature

In summary, a 50-g piece of ice at 0.0C is placed into a mixture of 200 g of water and an 80-g aluminum container at 30C. The final equilibrium temperature is found by setting the heat gained by the ice and water equal to the heat lost by the water and container, and solving for the final temperature. Using the specific heat for aluminum (900 J/Kg*C), the specific heat of water (4186 J/Kg*C), and the latent heat of fusion (3.33 X10^5), the final equilibrium temperature is found to be 9.5 Celcius.
  • #1
raymondmorrow
3
0

Homework Statement



A 50-g of ice, initially at 0.0C, is dropped into 200 g of water in an 80-g
aluminum container, both initially at 30C. What is the final equilibrium
temperature? (Specific heat for aluminum is 900 J/Kg*C, the specific heat of
water is 4186 J/Kg*C, and Lf=3.33 X10^5).

Homework Equations





The Attempt at a Solution



I am trying to help my wife with this. I have no clue what to do. Can you help me find somewhere it will show me what I need to do?
 
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  • #2
raymondmorrow said:

Homework Statement



A 50-g of ice, initially at 0.0C, is dropped into 200 g of water in an 80-g
aluminum container, both initially at 30C. What is the final equilibrium
temperature? (Specific heat for aluminum is 900 J/Kg*C, the specific heat of
water is 4186 J/Kg*C, and Lf=3.33 X10^5).

Homework Equations





The Attempt at a Solution



I am trying to help my wife with this. I have no clue what to do. Can you help me find somewhere it will show me what I need to do?
Welcome to PF raymondmorrow,

What are your wife's thoughts on the question? What has she tried already?
 
  • #3
heat lost by metal= M metal*C metal*change in temperature

heat gained by ice-water= (Q ice + Q water)+ m water*C water* change in temperature

heat lost by metal= heat gained by ice-water

I keep trying but I can't get the right answer (9.5 Celcius) Any help is very appreciated.
 
  • #4
raymondmorrow said:
heat gained by ice-water= (Q ice + Q water)+ m water*C water* change in temperature
Can you write this expression more explicitly? What are Q ice and Q water for example?
 
  • #5
Heat lost by mutual:
(0.08kg)(900 j/kg*C)(30C-T)
=72 J/C(30C-T)

Ice to melt:
Q=(M ice)(lf)
=(0.05)(3.33x10^5)
=16650J
0.05kg * 0.2kg = 0.25kg

heat gained by water from ice:
(16650) + (0.25)(4.19 x 10 ^ 3)(T-0C)
16650+1047.5(T--0C)
16650+1047.5T

72 J/C(30C-T)=16650+1047.5T
2160-72J/CT=16650+1047.5T
-72 J/CT=14490+1047.5T (after subtracting 2160 from both sides)
-1195T=14490 (After subtracting 1047.5T from both sides)
=1.29
 
  • #6
raymondmorrow said:
Heat lost by mutual:
(0.08kg)(900 j/kg*C)(30C-T)
=72 J/C(30C-T)

Ice to melt:
Q=(M ice)(lf)
=(0.05)(3.33x10^5)
=16650J
0.05kg * 0.2kg = 0.25kg

heat gained by water from ice:
(16650) + (0.25)(4.19 x 10 ^ 3)(T-0C)
16650+1047.5(T--0C)
16650+1047.5T

72 J/C(30C-T)=16650+1047.5T
2160-72J/CT=16650+1047.5T
-72 J/CT=14490+1047.5T (after subtracting 2160 from both sides)
-1195T=14490 (After subtracting 1047.5T from both sides)
=1.29

You can't add the masses of the two waters, since they are at different initial temperatures and the 50 g of water is taking heat away from the 200 g of water and the container. You need separate terms for each one. You should have 4 terms to work with:

the heat gained by the 50 g of ice at 0 degrees to become water at 0 degrees (Q1 let's say)
the heat gained by the 50 g of water now at 0 degrees to get to Tf (Q2)
the heat lost by the 200 g of water in the container to get to Tf (Q3)
the heat lost by the aluminum container to get to Tf (Q4)

Your equation should be laid out like

heat used to melt the ice (Q1) and warm the 50 g of water (Q2) = heat lost by 200 g of water (Q3) and heat lost by container (Q4).

Put the terms in the right place and solve for Tf.
 

1. What is the relationship between water, ice, and metal in terms of temperature?

The relationship between water, ice, and metal in terms of temperature is that they all have specific temperature ranges at which they exist in different states. Water can exist as a liquid, solid (ice), or gas (water vapor) depending on its temperature. Metal, on the other hand, does not change state but its temperature can affect its properties such as its strength and conductivity.

2. How does temperature affect the properties of water, ice, and metal?

Temperature greatly affects the properties of water, ice, and metal. For water, temperature determines whether it is liquid, solid, or gas. It also affects its density, heat capacity, and ability to dissolve other substances. In the case of ice, temperature affects its stability and ability to exist in its solid form. For metal, temperature affects its strength, ductility, and electrical conductivity.

3. How can temperature be measured for water, ice, and metal?

Temperature can be measured using various instruments such as thermometers, infrared cameras, and thermocouples. For water and ice, a thermometer is commonly used to measure their temperature. For metal, a thermocouple is often used as it can withstand high temperatures and provide accurate readings.

4. Are there any factors that can influence the temperature of water, ice, and metal?

Yes, there are many factors that can influence the temperature of water, ice, and metal. These include the surrounding environment, such as air temperature and pressure, as well as the amount of heat energy applied or removed from the substances. Other factors like impurities, motion, and chemical reactions can also affect temperature.

5. How is the concept of specific heat related to water, ice, and metal temperature?

Specific heat is the amount of heat energy required to raise the temperature of a substance by a certain amount. Water has a higher specific heat compared to ice and metal, meaning it requires more energy to change its temperature. This is why water is often used to regulate and maintain temperature in systems, such as in heating and cooling systems. Metal, on the other hand, has a lower specific heat, making it easier to heat up or cool down compared to water and ice.

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