Respuesta :
Answer:
[tex]T_{2,H_2O}=89^oC[/tex]
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:
[tex]Q_{1,H_2O}=-Q_{2,H_2O}[/tex]
Which in terms of masses, heat capacities and temperatures is:
[tex]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})[/tex]
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:
[tex]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}=\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{-m_{2,H_2O}} \\\\T_{2,H_2O}=T_{EQ}+\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{m_{2,H_2O}}[/tex]
With the given data we obtain:
[tex]T_{2,H_2O}=T_{EQ}+\frac{m_{1,H_2O}*(T_{EQ}-T_{1,H_2O})}{m_{2,H_2O}}\\\\T_{2,H_2O}=48.9^oC+\frac{102g*(48.9^oC-22.6^oC)}{66.9g} =48.9^oC+40.1^oC\\T_{2,H_2O}=89^oC[/tex]
Which means the second sample was hot.
Regards.
Answer:
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
