Question

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

Answer 1

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

Step-by-step 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

Answer 2

`T_(2,H_2O)=89^oC`

Step-by-step explanation:

Hello,

In this case, we notice that the energy gained by the first sample of water is lost by the second sample of water as they are heated and cooled respectively, therefore, in terms of heats:

`Q_(1,H_2O)=-Q_(2,H_2O)`

Which in terms of masses, heat capacities and temperatures is:

`m_(1,H_2O)*Cp_(1,H_2O)*(T_(EQ)-T_(1,H_2O))=-m_(2,H_2O)*Cp_(2,H_2O)*(T_(EQ)-T_(2,H_2O))`

In such a way, as the heat capacity is the same and the initial temperature of the cold water is required, we solve for it via:

`m_(1,H_2O)*(T_(EQ)-T_(1,H_2O))=-m_(2,H_2O)*(T_(EQ)-T_(2,H_2O))\\T_(EQ)-T_(2,H_2O)=(m_(1,H_2O)*(T_(EQ)-T_(1,H_2O)))/(-m_(2,H_2O)) \\\\T_(2,H_2O)=T_(EQ)+(m_(1,H_2O)*(T_(EQ)-T_(1,H_2O)))/(m_(2,H_2O))`

With the given data we obtain:

`T_(2,H_2O)=T_(EQ)+(m_(1,H_2O)*(T_(EQ)-T_(1,H_2O)))/(m_(2,H_2O))\\\\T_(2,H_2O)=48.9^oC+(102g*(48.9^oC-22.6^oC))/(66.9g) =48.9^oC+40.1^oC\\T_(2,H_2O)=89^oC`

Which means the second sample was hot.

Regards.