Water (c= 4180 j/kg. °c) is to be heated by solar heated hot air (cp = 1010 j/kg•°c) in a double-pipe counterflow heat exchanger. air enters the heat exchanger at 90°c at a rate of 0.3 kg/s, while water enters at 22°c at a rate of 0.1 kg/s. the overall heat transfer coefficient based on the inner side of the tube is given to be 80 w/m.°c. the length of the tube is 12 m and the internal diameter of the tube is 1.2 cm. determine the outlet temperatures of the water and the air.( 27.3°c, 82.7°c)

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

since electrons are negative the answer is neagtaive

the answer is a for sure mark brainliest i had got this right on usatestprep

answer///carbon;