Em uma garrafa térmica, são colocados 200 g de água à temperatura de 30 ºC e uma pedra de gelo de 50 g, à temperatura de –10 ºC. Após o equilíbrio térmico,

Let's break down the problem step by step. We have 200g of water at 30°C and 50g of ice at -10°C in a thermally insulated bottle. After thermal equilibrium is reached, we need to determine the final temperature of the system.

The key to this problem is to understand that the heat gained by the ice (which will melt and then warm up) is equal to the heat lost by the water (which will cool down). Let's calculate the heat gained by the ice:

First, the ice will melt, absorbing heat in the process. The heat of fusion of water is approximately 334 J/g, so the heat required to melt 50g of ice is:

Q_fus = 50g × 334 J/g = 16700 J

Once the ice has melted, the resulting water will warm up to the final temperature. Let's assume the final temperature is T (in °C). The heat gained by the water is:

Q_warm = 50g × c × (T - 0°C)

where c is the specific heat capacity of water, approximately 4.184 J/g°C.

The heat lost by the water is:

Q_cool = 200g × c × (30°C - T)

Since the system is thermally insulated, the heat gained by the ice and water is equal to the heat lost by the water:

Q_fus + Q_warm = Q_cool

Substituting the expressions, we get:

16700 J + 50g × c × (T - 0°C) = 200g × c × (30°C - T)

Solving for T, we get:

T ≈ 7°C

Therefore, the correct answer is A) The ice completely melts, and the temperature of the system at equilibrium is approximately 7°C.

This result makes sense, as the heat gained by the ice (16700 J) is less than the heat lost by the water (approximately 25100 J), so the ice will completely melt, and the final temperature will be higher than 0°C but lower than 30°C.