How much boiling water at 100 ∘c must you add to this beaker so that the final temperature of the mixture will be 77 ∘c?

I'll discuss first the principle used behind the calculations that we're going to do later on. According to the Law of Conservation of Energy, energy is neither created nor destroyed. They are just transferred from one system to another. In this case, it is transferred through heat energy.

Now, you should know that there are two types of heat energy: latent heat and sensible heat. Latent heat is the heat you add or remove at a constant temperature when there is a phase change occurring. On the other hand, sensible heat is the heat you add or remove as you raise or reduce the temperature of a system. For a boiling water, a phase change occurs, so we're going to use latent heat for that one. The heat used to raise the temperature to 77°C is the sensible heat.

The equation for latent hear is mΔH, where ΔH for phase change from liquid to gas is called latent heat of vaporization. For water, ΔH = 334 Joules/gram. The equation for sensible heat is mCpΔT, where Cp is the specific heat of a substance. For water, Cp = 4.186 J/g-°C. 

However, we need to know what's inside the beaker initially. But there is no given data for this. Let's just assume that there is 100 grams of water at 25°C in the beaker initially. Therefore, our working equation would be:

mCpΔT = mΔH

We equate them because, as we said, heat is just transferred between the two. So, the energy is just equal. Substituting the values:

(10 g)(4.186 J/g-°C)(77°C - 25°C) = m(334 J/g)

Solving for m,
m = 6.517 g

Therefore, if there is initially 10 grams at 25°C of water in the beaker, then you would need 6.517 g of boiling water to be able to raise the temperature to 77°C.

The answer is 65.3312