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
