Calculate the amount of heat required to convert 155 g of ice at -10oC to steam at 110oC?
(155g)x[0-(-10)]C x (.50cal/g x C) = 775cal
(155g) x (80cal/g) = 12400cal
(155g)x(100-0)C x (1.00cal/g x C) = 15500cal
(155g) x (540cal/g) = 83700cal
775cal + 12400cal + 15500cal +83700cal = 112375cal
Is that correct?
1 answer
I don't know what the C stands for but you seem to have ignored it in the calculation. I didn't go through the calculations with a calculator but the methods seems OK EXCEPT that you haven't added in the heat from steam at 100 to steam at 110
5 answers