A 60.13 gram sample of iron (with a heat capacity of 0.450 J/gºC) is heated to 100.0 ºC. It is then transferred to a coffee cup calorimeter containing 46.92 g of water (specific heat of 4.184 J/ gºC) initially at 20.63 ºC. If the final temperature of the system is 23.59ºC, how much heat was absorbed by the bomb calorimeter?

SU

Explanation:

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

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.