sebuah lemari es memiliki koefisien performansi 5

Pembahasan

k subscript p equals fraction numerator T subscript r e n d a h end subscript over denominator T subscript t i n g g i end subscript minus T subscript r e n d a h end subscript end fraction

Maka

k subscript p. T subscript t i n g g i end subscript minus k subscript p. T subscript r e n d a h end subscript equals T subscript r e n d a h end subscript k subscript p. T subscript t i n g g i end subscript space equals space T subscript r e n d a h end subscript space plus space k subscript p. T subscript r e n d a h end subscript k subscript p. T subscript t i n g g i end subscript equals space T subscript r e n d a h end subscript space left parenthesis 1 plus space k subscript p right parenthesis T subscript r e n d a h end subscript equals space fraction numerator k subscript p. T subscript t i n g g i end subscript over denominator space left parenthesis 1 plus space k subscript p right parenthesis end fraction T subscript r e n d a h end subscript equals space fraction numerator open parentheses 5 close parentheses left parenthesis 300 right parenthesis over denominator space left parenthesis 1 plus space 5 right parenthesis end fraction T subscript r e n d a h end subscript equals space fraction numerator 1500 over denominator space 6 end fraction equals space 250 space K space equals space 250 space minus space 273 space equals space minus 23 to the power of omicron C

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