Apan containing 30 grams of water was allowed to cool from a temperature of 90.0 °c. if the amount of heat released is 1,500 joules, what is the approximate final temperature of the water? 76 °c 78 °c 81 °c 82 °c

Option B, 78.05 °C is the approximate final temperature of the water

Explanation:

Heat released

Where "m" stands for mass of any substance

"c " stands for specific heat of water  is equal to 4.186 joule/gram °C

stands for final temperature

stands for initial temperature

Substituting the available values in above equation, we get -

78.05 °C is the approximate final temperature of the water

Thus, option B is correct.

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