What is the energy change of a 6.05 g sample of liquid water that cools from 90°C to 30°C? Cp = 4.184 J/gºC

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

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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

answer seismograph;