PLEASE HEL! HOW DO I DO THIS??? 1. Calculate the amount of heat released when a 225 g sample of gold cools from 86.0oC to 32.5oC. The specific heat of gold is 0.129 J/goC

2. A 19.6 g sample of a metal was heated to 61.67oC. When it was placed into 26.7 g of water in a constant pressure calorimeter, the temperature of the water rose from 25.00oC to 30.00oC. Determine the specific heat of the metal if the specific heat of H2O(liq) is 4.18 J/goC. Ignore any contribution from the container.

1552.83J Released

Explanation:

1. mass/m=225

Initial temp:86C, final:32.5C

Changed Temp: 32.5-86= -53.5C

s=0.129 J/gC

Formula: q= m times s times changed Temp.

q=(225)(0.129)(-53.5)

q= -1552.83 J

q=1552.83 J Released

the answer is c i believe

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