the sum of heats gained is zero.
Heatgainedicemelting+heatgainedwater=0
massice*Hf+53.4grams*specificheatwater*(0-65)=0
massice=53.4g*1cal/gramC *53.4C / 80cal/g
= 53.4*65/80 grams
How much ice (in grams) would have to melt to lower the temperature of 53.4g of water from 65 C to 0C?
I know there are more than one step here, I would appreciate an answer that does not do the work for me, but show me how to lay out the problems and in which order... thank you!
4 answers
Thank you Bob for your time! I believe there was a misunderstanding in the question,my mistake... C is equal to degree celcius.
Note the correct spelling of celsius.
Celsius is all that is needed. It need not be converted to kelvin.
Celsius is all that is needed. It need not be converted to kelvin.
Thank you again!