When 102 g of water at a temperature of 22.6 °C is mixed with 66.9 g of water at an unknown temperature, the final temperature of the resulting mixture is 48.9 °C. What was the initial temperature of the second sample of water? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K.) Initial temperature = °C

The initial temperature of the second sample of water is 8.8 °C

Explanation:

Step 1 :Data given

Mass of water = 102 grams

Initial temperature of water = 22.6 °C =

Mass of other water = 66.9 grams

The final temperature is 48.9 °C

The specific heat capacity of liquid water is 4.184 J/g *K = 4.184 J/g°C

Step 2: Calculate the initial temperature of the second sample water

Heat gained = heat lost

Qgained = -Q lost

Q = m* C* ΔT

m(water1)*C(water)*ΔT(water1) = -m(water2) * C(water) *ΔT(water2)

⇒with m(water1) = the mass of the water at 22.6 °C = 102 grams

⇒with C(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT = the change of water of the first sample = 48.9 - 22.6 = 26.3 °C

⇒with m(water2) = the mass of the other unknown sample water = 66.9 grams

⇒with with C(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT = the change of water of the second sample = 48.9 - T1

102 * 4.184 * 26.3 °C = 66.9 * 4.184 * (48.9 - T1)

102 * 26.3 = 66.9 * (48.9 - T1)

2682.6 = 3271.4 - 66.9 T1

-588.8 = -66.9 T1

T1 = 8.8 °C

The initial temperature of the second sample of water is 8.8 °C

it would be a gas formation

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