Answer : The final temperature is, ![25.0^oC](https://tex.z-dn.net/?f=25.0%5EoC)
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
![q_1=-q_2](https://tex.z-dn.net/?f=q_1%3D-q_2)
![m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_f-T_1%29%3D-m_2%5Ctimes%20c_2%5Ctimes%20%28T_f-T_2%29)
where,
= specific heat of ice = ![2.09J/g^oC](https://tex.z-dn.net/?f=2.09J%2Fg%5EoC)
= specific heat of water = ![4.18J/g^oC](https://tex.z-dn.net/?f=4.18J%2Fg%5EoC)
= mass of ice = 50 g
= mass of water = 200 g
= final temperature = ?
= initial temperature of ice = ![-15^oC](https://tex.z-dn.net/?f=-15%5EoC)
= initial temperature of water = ![30^oC](https://tex.z-dn.net/?f=30%5EoC)
Now put all the given values in the above formula, we get:
![50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC](https://tex.z-dn.net/?f=50g%5Ctimes%202.09J%2Fg%5EoC%5Ctimes%20%28T_f-%28-15%29%29%5EoC%3D-200g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_f-30%29%5EoC)
![T_f=25.0^oC](https://tex.z-dn.net/?f=T_f%3D25.0%5EoC)
Therefore, the final temperature is, ![25.0^oC](https://tex.z-dn.net/?f=25.0%5EoC)
Answer:
c > √(2ab)
Explanation:
In this exercise we are asked to find the condition for c in such a way that the results have been real
The given equation is
½ a t² - c t + b = 0
we can see that this is a quadratic equation whose solution is
t = [c ±√(c² - 4 (½ a) b)] / 2
for the results to be real, the square root must be real, so the radicand must be greater than zero
c² -2a b > 0
c > √(2ab)