Por um resistor com resistência constante igual a 1000 ohms circula uma corrente de 2 ampères durante 5 minutos. Se esse resistor está imerso em um recipiente adiabático de capacidade térmica igual a 300 J/0C contendo 5 kg de água, inicialmente a 10° C, desprezarmos as perdas para o ambiente e considerarmos o calor específico da água igual a 4,17 J/g.0C, a temperatura final atingida pela água será igual a

Let's break down the problem step by step to find the correct answer.

First, we need to calculate the energy dissipated by the resistor. Since the resistor has a constant resistance of 1000 ohms, we can use Joule's law to find the energy dissipated:

Q = R * I^2 * t

where Q is the energy dissipated, R is the resistance, I is the current, and t is the time. Plugging in the values given, we get:

Q = 1000 ohms * (2 A)^2 * 5 minutes = 1200 J

Now, we need to find the temperature change of the water. We can use the formula:

ΔT = Q / (m * c)

where ΔT is the temperature change, m is the mass of the water, and c is the specific heat capacity of water. Plugging in the values given, we get:

ΔT = 1200 J / (5 kg * 4.17 J/g°C) = 56.7°C

Since the initial temperature of the water is 10°C, the final temperature will be:

T_f = 10°C + 56.7°C = 66.7°C

Therefore, the correct answer is:

C) 66.7°C

This answer makes sense because the energy dissipated by the resistor is transferred to the water, causing its temperature to increase. The final temperature of the water is higher than the initial temperature, which is consistent with the laws of thermodynamics.