Uma pessoa precisa de 1 L de água a 50ºC para fazer uma compressa de água quente. Para isso, ela resolve misturar, no interior de uma garrafa térmica de capacidade térmica desprezível, uma certa quantidade de água à temperatura ambiente de 20ºC com uma outra quantidade de água a 100ºC que ela acabara de ferver. Assinale a alternativa que corresponde às quantidades de água a 20ºC e 100ºC que devem ser misturadas para que a pessoa consiga atingir seu objetivo.

Let's break down the problem step by step. We want to find the quantities of water at 20°C and 100°C that need to be mixed to obtain 1 L of water at 50°C.

First, let's define the quantities of water at 20°C and 100°C as x and y, respectively. We know that the total quantity of water is 1 L, so we can write an equation:

x + y = 1000 g

Since the final temperature of the mixture is 50°C, we can use the principle of heat transfer to write another equation:

(x × 20°C) + (y × 100°C) = (1000 g × 50°C)

Simplifying this equation, we get:

20x + 100y = 50000

Now, we can solve this system of equations using substitution or elimination. Let's use substitution. Rearranging the first equation, we get:

y = 1000 - x

Substituting this into the second equation, we get:

20x + 100(1000 - x) = 50000

Simplifying and solving for x, we get:

x = 625 g

Now, we can find y:

y = 1000 - 625 = 375 g

Therefore, the correct answer is alternative D) 625 g at 20°C and 375 g at 100°C.

This solution makes sense because we need to mix a larger quantity of cooler water with a smaller quantity of hotter water to obtain the desired temperature.