How many joules are needed to heat 8.50 grams of ice from -10.0 degrees C to 
25 degrees C?

Hi teatea!!

the following equations will be used: 
For the change of the temperature of ice and liquid water without
changing phase:
Heat[cal]= (mass[g]) x (change of temperature[ºC]) x (specific
heat[cal/ºC.g])

For the change of phase from ice to liquid water: 
Heat{cal] = (mass[g]) x (Heat of fusion[cal/g]) 

This is a 3-step problem:

Step 1: Calculate the heat for change the temperature of 8.5g of ice
from -10ºC to 0ºC.

Specific heat of ice = 0.5 cal/ºC.g

Heat1 = 8.5g x 10ºC x 0.5 cal/ºC.g = 42.5 cal


Step 2: Calculate the heat needed to melt the ice:

Heat of fusion for water = 80 cal/g

Heat2 = 8.5g x 80 cal/g = 680 cal


Step 3: Calculate the heat for change the temperature of 8.5g of water
from 0ºC to 25ºC.

Specific heat of water = 1 cal/ºC.g

Heat3 = 8.5g x 25ºC x 1 cal/ºC.g = 212.5 cal


Step4: Calculate the total heat.

Total Heat = Heat1 + Heat2 + Heat3 = 42.5 cal + 680 cal + 212.5 cal =
935 cal


I hope this helps you.

Best regards.
livioflores-ga

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Clarification of Answer by livioflores-ga on 15 Jul 2003 01:11 PDT

Ooops!! I forgot the conversion from calories to Joules!!

The conversion is 1 cal = 4.184 J.

we have a Total Heat of 935 cal; 
this heat is equivalent to (4.184 J/cal * 935 cal) = 3912.04 J

Regards.
livioflores-ga

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Clarification of Answer by livioflores-ga on 15 Jul 2003 01:26 PDT

Hello!!

A brief clarification:
In some places the values for Specific Heat varies a little from the
used for me to answer this question.
For example if we consider the Specific Heat of ice equal to 0.478
cal/ºC.g ; Heat1 = 8.5g x 10ºC x 0.478 cal/ºC.g = 40.63 cal ;
Then
Total Heat = 40.63 cal + 680 cal + 212.5 cal = 933.13 cal = 3904.216
J.

See this page:
"Specific Heats":
http://www.austin.cc.tx.us/rvsmthsc/chem/chem-Heat.html

Sincerely.
livioflores-ga