Cool Coffee with Melting Ice Cube: Solve for T

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Homework Help Overview

The problem involves a thermal equilibrium scenario where hot coffee cools down after an ice cube melts within it. The context is centered around heat transfer, specifically involving the specific heat of water and the latent heat of fusion.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to set up an equation based on energy conservation but indicates that their solution did not yield the expected results. Some participants question the use of specific values and units in the calculations.

Discussion Status

The discussion is ongoing, with participants providing feedback on the equation's structure and unit considerations. There is no explicit consensus yet, but some guidance has been offered regarding the equation's validity.

Contextual Notes

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Homework Statement


An insulated Thermos contains 190 cm3 of hot coffee at 83.0°C. You put in a 18.0 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 and equilibrium is reached? 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.

(the ice is initially at 0 C)


The Attempt at a Solution


i tried this but didnt work :
190 * 4.186 * (83.0 - T) = 18.0 * 333 + 18.0 * 4.186 * (T -0) and solved for T.

can someone help please?
 
Last edited:
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Why 3333?

Pay attention to units.
 
Borek said:
Why 3333?

Pay attention to units.

i edited my previous post because i wrote 3333 here by mistake..

btw i tried both 333 and .333 and didnt work
 
Other than that equation looks OK to me.
 

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