3690 j of heat are added to a 21.5 g sample

of iron at 30◦c. what is the final temperature of the iron? the specific heat of iron is

0.449 j/g · k.

answer in units of k.

been stuck on this for four days thxs

693K

Explanation:

The enthalpy change in the iron is 3690J

We now apply the formula for enthalpy change which is ΔH=mC∅ where ∅ is the temperature change, m the mass of the substance, and C the specific heat capacity for the substance.

ΔH in this case is 3690J.

Therefore 3690J=21.5g×0.449J/g.K×∅

as we are looking for ∅, we make it the subject of the formula.

∅=3690J/(21.5g×0.44J/g)

∅=390

Temperature=30°C +390

=420°+273

=693K

pie chart because it is in percents ( % )

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