What is the specific heat of a substance that absorbs 2500 joules of heat when a sample of 100g of the substance increases in temperature from 10.°C to 70.°

0.417 J/gºC

Explanation:

From the question given above, the following data were obtained:

Heat (Q) absorbed = 2500 J

Mass (M) of substance = 100 g

Initial temperature (T1) = 10 °C

Final temperature (T2) = 70 °C

Specific heat capacity (C) =?

Next, we shall determine the change in temperature (ΔT). This can be obtained as follow:

Initial temperature (T1) = 10 °C

Final temperature (T2) = 70 °C

Change in temperature (ΔT) =?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 70 – 10

Change in temperature (ΔT) = 60 °C

Finally, we shall determine the specific heat capacity of the substance as follow:

Heat (Q) absorbed = 2500 J

Mass (M) of substance = 100 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) =?

Q = MCΔT

2500 = 100 × C × 60

2500 = 6000 × C

Divide both side by 6000

C = 2500 / 6000

C = 0.417 J/gºC

Therefore, the specific heat capacity of substance is 0.417 J/gºC

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