Dois recipientes de material termicamente isolante contêm cada um l0g de água a 0ºC. Deseja-se aquecer até uma mêsma temperatura os conteúdos dos dois recipientes, mas sem misturá-los. Para isso é usado um bloco de 100g de uma liga metálica inicialmente à temperatura de 90%. O bloco é imerso durante um certo tempo num dos recipientes e depois transferido para o outro, nele permanecendo até ser atingido o equilíbrio térmico. O calor específico da água é dez vezes maior que o da liga. A temperatura do bloco, por ocasião da transferência, deve então ser igual a

Let's break down the problem step by step.

We have two containers with 10g of water each at 0°C. We want to heat them up to the same temperature without mixing them. To do this, we use a 100g metal block initially at 90°C. The block is immersed in one of the containers for a certain time and then transferred to the other, where it remains until thermal equilibrium is reached.

We know that the specific heat of water is 10 times greater than that of the metal block. This information is crucial to solve the problem.

Let's analyze the situation:

Initially, the metal block is at 90°C, and the water in both containers is at 0°C. When the block is immersed in one of the containers, heat will be transferred from the block to the water. Since the specific heat of water is higher, the temperature of the block will decrease more rapidly than the temperature of the water.

When the block reaches thermal equilibrium with the water in the first container, it will have lost some heat, and its temperature will be lower than 90°C. Let's call this temperature T.

Now, the block is transferred to the second container. Since the block is at a higher temperature than the water in the second container, heat will be transferred from the block to the water again.

At this point, we can apply the principle of energy conservation. The heat lost by the block as it cools down from 90°C to T is equal to the heat gained by the water in both containers as they warm up from 0°C to T.

Using the concept of specific heat, we can set up an equation to relate the heat lost by the block to the heat gained by the water. Let's denote the specific heat of the metal block as c_m and that of water as c_w.

Since the specific heat of water is 10 times greater than that of the metal block, we can write:

c_w = 10c_m

The heat lost by the block as it cools down from 90°C to T is:

Q_lost = m_blockc_m(90°C - T)

The heat gained by the water in both containers as they warm up from 0°C to T is:

Q_gained = 2m_waterc_w(T - 0°C)

Since energy is conserved, we can set up the equation:

Q_lost = Q_gained

Substituting the expressions for Q_lost and Q_gained, we get:

m_blockc_m(90°C - T) = 2m_waterc_w(T - 0°C)

Simplifying the equation and solving for T, we get:

T = 60°C

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

This problem requires a deep understanding of thermodynamic principles, including energy conservation and specific heat. By applying these concepts, we can solve the problem step by step and arrive at the correct answer.