Respuesta :
Answer:
65.1 °C
Explanation:
m = mass of the water = 0.5 kg
[tex]T_{i}[/tex] = initial temperature of water = 35.00 °C
[tex]T_{f}[/tex] = final temperature of water
c = specific heat of water = 4186 J/(kg °C)
Q = Amount of heat removed from water = 6.3 x 10⁴ J
Amount of heat removed from water is given as
Q = m c ([tex]T_{f}[/tex] - [tex]T_{i}[/tex])
Inserting the values
6.3 x 10⁴ = (0.5) (4186) ([tex]T_{f}[/tex] - 35.00)
[tex]T_{f}[/tex] = 65.1 °C
Answer : The final temperature of the water is [tex]65.10^oC[/tex]
Explanation :
Formula used :
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat released = [tex]6.300\times 10^{4}J[/tex]
m = mass of water = 0.5000 kg
c = specific heat of water = [tex]4186J/kg^oC[/tex]
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_{initial}[/tex] = initial temperature = [tex]35.00^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]6.300\times 10^{4}J=(0.5000kg)\times (4186J/kg^oC)\times (T_{final}-35.00)^oC[/tex]
[tex]T_{final}=65.10^oC[/tex]
Therefore, the final temperature of the water is [tex]65.10^oC[/tex]
