Asked by Jose
How many grams of ice must be added to 100 grams of water at 60 degrees so that its temperature drops to 20 degrees celcius?
As far as i know I have to use:
(Mass)*(specific heat of water) *(finaltemperature - initial temperature = ??
As far as i know I have to use:
(Mass)*(specific heat of water) *(finaltemperature - initial temperature = ??
Answers
Answered by
bobpursley
What is the temperature of the ice when you put it in?
heat to lose=100*specifheatwater*40
heat absorbed by ice: mass*specificheatice*(Ti)+ mass*Heatfusionice*(20)
set them equal, after putting in Tinitial of ice, and solve for the mass of ice.
heat to lose=100*specifheatwater*40
heat absorbed by ice: mass*specificheatice*(Ti)+ mass*Heatfusionice*(20)
set them equal, after putting in Tinitial of ice, and solve for the mass of ice.
Answered by
Jose
the temperature of ice according to our physics teacher is -5 degrees celcius
Answered by
bobpursley
Then
massice*(specificheatice)5+massice*heatfusionice*20= 100*spcificheatdwater*40
massice*(specificheatice)5+massice*heatfusionice*20= 100*spcificheatdwater*40
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