Respuesta :
Answer:
63.52°C is the final temperature
Explanation:
1) 131.0 mL of water at 26.0°C
Mass of water = m
Volume of the water =131.0 mL
Density of the water = 1.00 g/mL
[tex]Density=1.00 g/mL=\frac{m}{131.0 mL}[/tex]
m = 131.0 g
Initial temperature of the water = [tex]T_i[/tex] = 26.0°C
Final temperature of the water = [tex]T_f[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i[/tex]
Heat absorbed 131.0 g of water = Q
[tex]Q=m\times c\times \Delta T[/tex]
2) 81.0 mL of water at 85.0°C
Mass of water = m'
Volume of the water =81.0 mL
Density of the water = 1.00 g/mL
[tex]Density=1.00 g/mL=\frac{m'}{81.0 mL}[/tex]
m' = 81.0 g
Initial temperature of the water = [tex]T_i'[/tex] = 85.0°C
Final temperature of the water = [tex]T_f'[/tex]
Change in temperature ,[tex]\Delta T'=T_f'-T_i'[/tex]
Heat lost by 81.0 g of water = Q'
[tex]Q'=m'\times c\times \Delta T'[/tex]
After mixing both liquids the final temperature will become equal fro both liquids.
[tex]T_f=T_f'[/tex]
Since, heat lost by the water at higher temperature will be equal to heat absorbed by the water at lower temperature.
Q=-Q' (Law of conservation of energy.)
Let the specific heat of water be c
[tex]m\times c\times \Delta T=m'\times c\times \Delta T'[/tex]
[tex]131.0 g\times c(T_f-26^oC)=-(81.0.0 g\times c(T_f-85^oC))[/tex]
[tex]T_f=63.52^oC[/tex]
63.52°C is the final temperature
According to conservation law, the amount of energy in a closed and isolated system remains constant and does not get reduced or added. The final temperature of the system is 63.52°C.
What are mass and temperature?
First, calculate the mass of the 131 mL of water:
Given,
- Mass (m)= ?
- Volume (V)= 131.0 mL
- Density = 1.00 g/ml
[tex]\begin{aligned}\rm Mass &= \rm Volume \times \rm density\\\\\rm m &= 131.0 \;\rm g\end{aligned}[/tex]
Given,
Initial temperature [tex]\rm (T_{i})[/tex] = 26.0°C
The final temperature of the water = [tex]\rm (T_{f})[/tex]
Change in the temperature is calculated as,
[tex]\rm \Delta T = T_{f} - T_{i}[/tex]
Heat absorbed by 131.0 g of water = Q
The formula to calculate heat absorbed is,
[tex]\rm Q = m \times c \times \Delta T[/tex]
Second, calculate the mass of the 81 mL of water:
Given,
- Mass (m') = ?
- Volume (V')= 81.0 mL
- Density = 1.00 g/ml
[tex]\begin{aligned}\rm Mass' &= \rm Volume \times \rm density\\\\\rm m' &= 81.0 \;\rm g\end{aligned}[/tex]
Given,
Initial temperature [tex]\rm (T'_{i})[/tex] = 26.0°C
The final temperature of the water = [tex]\rm (T'_{f})[/tex]
Change in the temperature is calculated as,
[tex]\rm \Delta T' = T'_{f} - T'_{i}[/tex]
Heat lost by 81.0 g of water = Q'
The formula to calculate heat lost is,
[tex]\rm Q' = m' \times c \times \Delta T'[/tex]
Both the liquids are mixed and the final temperature will be equivalent and given as, [tex]\Delta \rm T_{f} = \Delta T'_{f}[/tex]
According to the law of conservation,
Heat lost by the water (Q') = Heat absorbed by the water (-Q)
[tex]\begin{aligned}\rm m \times c \times \Delta T &= \rm m' \times c \times \Delta T'\\\\131 \times\rm c(T_{f} -26) &= -(81 \times\rm c(T_{f}- 85))\\\\\rm T_{f} &= 63.52 \circC\end{aligned}[/tex]
Therefore, 63.52°C is the final temperature.
Learn more about the law of conservation here:
https://brainly.com/question/2284197
