5,500 crates of apples are in storage at 3'C. An addtional 500 crates enter the storage cooler @ a temp of 29'C and are chilled to storage temp in 24 hours. The ave.mass of apples per crate is 27kg & the crate has a mass of 4.5kg & a specific heat of 2.5 kj/kg-k. det. the product load in kw.
the book says 16.083 kw but i cant get to this answer.
please show me the process..
2007-09-22 00:40:17 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Engineering
the specific heat of apples is not given..
2007-09-22 01:01:35 · update #1
17.014 kw is the closest I could get, but my specific heat for apples was assumed to be 0.9 btu/lb-deg f = 3.76812 kj/kg-k
and this might be where the difference is.
500 apple crates x 27 kg of apples = 13,500 kg apples
500 apple crates x 4.5 kg of wood = 2,250 kg wood
Q = m x cp x (t2-t1)
Q apples = 13,500kg x 3.76812 kj/kg-k x 26 deg k = 1,322,610 kj
Q wood = 2,250kg x 2.5kj/kg-k x 26 deg k = 146,250 kj
Q total = 1,322,610kj + 146,250kj = 1,468,860 kj
1,468.860 kj x .000278 kw-hr/kj = 408.34 kw-hr
408.34 kw-hr/24 hr = 17.014 kw
2007-09-22 17:37:28 · answer #1 · answered by gatorbait 7 · 0⤊ 0⤋
You have forgotten to mention the specific heat of the apple.s
2007-09-22 07:58:04 · answer #2 · answered by Pandian p.c. 3 · 0⤊ 0⤋