9 A 75 gram sample of water is initially at 30 degrees Celsius. What would it's temperature be if
it absorbed 2000 J of heat?

T2 = 36.38°C

Explanation:

Given data:

Mass of water = 75 g

Initial temperature = 30 °C

Final temperature = ?

Heat absorbed = 2000 J

Solution:

Specific heat capacity of water is 4.18 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

2000 J = 75 g×4.18 J/g.°C × T2- T1

2000 J = 313.5 J/°C × T2- T1

2000 J = 313.5 J/°C × T2 - 30 °C

2000 J / 313.5 J/°C  = T2 - 30 °C

6.38 °C = T2 - 30 °C

T2 = 6.38 °C +  30°C

T2 = 36.38°C

the ratio of the conjugate base to the acid required to attain the desired ph.