An Ice Cube is Added to a Thermos of Coffee

In summary, the energy used in melting the ice cube was 4.995 kJ and the energy released as the coffee cooled to 87 degrees was 273.15 kJ.
  • #1
justine411
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Homework Statement



An insulated Thermos contains 110 cm3 of hot coffee at 87.0°C. You put in a 15.0 g ice cube at its melting point to cool the coffee. By how many degrees (in Celsius) has your coffee cooled once the ice has melted? Treat the coffee as though it were pure water and neglect energy exchanges with the environment. The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg. The density of water is 1.00 g/cm3.

Homework Equations



Q=(m)(c)(Tf-Ti)
Q=(L)(m)

The Attempt at a Solution



I tried a couple of things. I am confident I could do the question if I didn't have to deal with the latent heat of fusion. I made two equations:
q= (mass of cube)(specific heat)(273.15-Tf)
q= (mass of water)(specific heat)(Tf-360.15)
and for Q of melting, I got 4.995 kJ. I'm just not sure how to incorporate that in. I tried adding it to the specific heat in the first equation and I tried multiplying it in, but neither approach worked. Suggestions Please?
 
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  • #2
Sorry...it seems like some people have viewed, but haven't posted yet. Is there anything I did wrong or should add? Basically, I'm just trying to figure out how the latent heat of fusion is incorporated into the two equations I set up above.
 
  • #3
You have calculated the energy used in melting the ice cube (energy of fusion). Now you need to calculate two more things. What is the energy required to raise the 15g= 15cc ice water produced by the ice cube to "T" degrees (for unknown T)? What is the energy released as the 110 cc of coffee cools from 87 to T degrees? Both of those, of course, depend on T.

Energy of fusion+ energy to raise the icewater to T= energy released by cooling the coffee to T.

Solve that equation for T.
 
  • #4
Ok. I'm still not getting the right answer for some reason! I have for my expression (Lm)+mc(T-273.15)=-mc(360.15-T)
What am I missing now?
 

FAQ: An Ice Cube is Added to a Thermos of Coffee

1. What happens to the temperature of the coffee when an ice cube is added to a thermos?

When an ice cube is added to a thermos of coffee, the temperature of the coffee will decrease. This is because the ice cube will absorb some of the heat energy from the coffee as it melts, causing the overall temperature to decrease.

2. How does the thermos keep the coffee from getting cold?

The thermos is designed with a double-walled vacuum insulation, which helps to prevent heat transfer between the inside and outside of the thermos. This means that the heat from the coffee will be trapped inside, keeping it from getting cold quickly.

3. Will the ice cube eventually melt in the thermos?

Yes, the ice cube will eventually melt in the thermos. Even though the vacuum insulation helps to slow down heat transfer, it cannot completely stop it. Eventually, the ice cube will melt due to the difference in temperature between the coffee and the ice cube.

4. How does the addition of the ice cube affect the overall volume of the coffee in the thermos?

Adding an ice cube to the thermos will increase the overall volume of the coffee. This is because the ice cube will melt and add water to the coffee, increasing its volume. However, the change in volume will be minimal compared to the total volume of the thermos.

5. Can the process of adding an ice cube to a thermos be reversed?

Yes, the process of adding an ice cube to a thermos can be reversed. If the thermos is opened and the ice cube is removed, the coffee will begin to warm up again. This is because the heat energy from the coffee will no longer be absorbed by the ice cube, causing the temperature to increase.

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