Question

TAT, =25

Answer 1

Answer:

70.07 g  

Explanation:

Data Given:

Amount of Heat release = 2.20 x 10⁴ J

initial temperature = 100 °C

Final temperature = 25 °C

Cs of water = 4.186 J/g °C

mass of water = ?

Solution:

Formula used

            Q = Cs.m.ΔT

rearrange the above equation to calculate the mass of water sample

           m =  Q / Cs.ΔT  .... . . . . . (1)

Where:

Q = amount of heat

As heat is released therefore Q will be negative

Q = -2.20 x 10⁴ J

Cs = specific heat of water = 4.186 J/g °C

m = mass

ΔT = (t2 - t1) = Change in temperature

First we have to find ΔT

                  ΔT = (t2 - t1)

                  ΔT = (25 °C - 100°C )

                  ΔT = -75 °C

Put values in above equation 1

                m = - 2.20 x 10⁴ J / 4.186 J/g °C x -75 °C

                m =  - 2.20 x 10⁴ J / - 313.95 (J/g)

                m =  70.07 g                  

So mass of water = 70.07 g