Thermo Question

  • Thread starter suspenc3
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  • #1
suspenc3
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I have a question involving putting an ice cube in a thermos of coffee. I used [tex]c_{ice}m_{ice}(T_f-T_i) + c_{coffee}m_{coffee}(T_f-T_i)=0[/tex]. Is this right? If so wouldn't the temperatures of the ice remain constant until it is all gone?It says that the ice is at 0'C.
 
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Answers and Replies

  • #2
Noein
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The approach is correct, but there is a term that is missing in your equation: the heat needed to completely melt the ice cube.
 
  • #3
suspenc3
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How do I find this..If say the ice was at -10'C then i could do this by using Tf=0..But since it is already at its melting point, I am kinda confused
 
  • #4
suspenc3
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waiitt...Q=Lm?
 
  • #5
turdferguson
312
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Heat Lost = Heat Gained, the heat lost by the coffee melts the ice first (mL) and then raises it to a higher temperature (mcdeltaT)
 
  • #6
suspenc3
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So my equation is going to become [tex]L_Fm + c_{ice}m_{ice}(T_f-T_i) + c_{coffee}m_{coffee}(T_f-T_i)=0[/tex]? and then will [tex]c_{ice}m_{ice}(T_f-T_i)[/tex] become zero since the ice isn't going to change Temp until it has changed phases
 
  • #7
turdferguson
312
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It depends on the numbers. If the final temperature is above 0 degrees, then there was enough heat to both melt the ice and raise the temperature of the new water above 0 degrees

Something else to point out is that the problem wouldn't be simpler if the ice started out at -10 degrees. You would need to determine (in this order) the heat used to raise the temperature to 0, the heat to melt the ice, and the heat to raise the water temperature above 0
 

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