Para aplicarmos o princípio das trocas de calor, um professor no laboratório de Física do IFPA deu a seguinte orientação: Separem num béquer 1 kg de água e aqueçam até 80ºC. Em seguida separem 500g de água na temperatura ambiente e derramem no béquer. Nesse caso, considerando a temperatura ambiente 20ºC, o calor específico da água 1 cal/g.ºC e a capacidade térmica do béquer desprezível, obtemos 1,5kg de água a:

Let's break down the problem step by step. We have 1 kg of water at 80°C, and we add 500g of water at room temperature (20°C). We need to find the final temperature of the mixture.

First, let's calculate the heat energy transferred from the hot water to the cold water. The specific heat capacity of water is 1 cal/g°C, so the heat energy transferred is:

Q = mcΔT = 1 kg * 1 cal/g°C * (80°C - T_f)

where T_f is the final temperature of the mixture.

Since the cold water is at room temperature, it will absorb heat energy from the hot water. The heat energy absorbed by the cold water is:

Q = mcΔT = 0.5 kg * 1 cal/g°C * (T_f - 20°C)

Since the heat energy transferred from the hot water is equal to the heat energy absorbed by the cold water, we can set up the equation:

1 kg * 1 cal/g°C * (80°C - T_f) = 0.5 kg * 1 cal/g°C * (T_f - 20°C)

Simplifying the equation, we get:

80°C - T_f = 0.5 * (T_f - 20°C)

Solving for T_f, we get:

T_f = 60°C

Therefore, the correct answer is E) 60°C.

This result makes sense because the final temperature of the mixture is between the initial temperatures of the hot and cold water.