calculate heat absorbed 125 grams of water at 72.0 degrees Celsius is converted
ID: 927850 • Letter: C
Question
calculate heat absorbed 125 grams of water at 72.0 degrees Celsius is converted to steam at 186.4 degrees Celsius. Specific heat of water is 4.18 J/g °c. heat of vaporization of water is 4.07 x10 to the 4th J/mol. specific heat of steam is 2.02 J/g°c.
the answer I got was 1.14x10 3joules. the book answer is 3.19 x 10 5th Joules.
please walk me through each stepped (typed preferably, it's hard for me to read handwriting) .
my calculations were: simplified
(125 )(4.18)x72.0 °C= 376 J
(125)18.02)= 6.93 MOL H20
(6.93)(4.07× 10 4th)= 282 J
(125)(2.02)(186.4°c)=471 J
Explanation / Answer
Q absorbe d = m*s*DT + n*DHvap +m*s*DT
= 125*4.18*(100-72.0)+(125/18)*40.7*10^3+ 125*2.02*(186.4-100)
= 319084.9 joule
= 3.19*10^5 joule.
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