A32.2 g iron rod, initially at 21.9 c, is submerged into an unknown mass of water at 63.5 c. in an insulated container. the final temperature of the mixture upon reaching thermal equilibrium is 59 2 c what is the mass of the water? express your answer to two significant figures

Answer : The mass of the water in two significant figures is,

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

where,

= specific heat of iron metal =

= specific heat of water =

= mass of iron metal = 32.3 g

= mass of water = ?

= final temperature of mixture =

= initial temperature of iron metal =

= initial temperature of water =

Now put all the given values in the above formula, we get

Therefore, the mass of the water in two significant figures is,

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