According to the Law of Conservation of Energy, energy is neither created nor destroyed. They are just transferred from one system to another. To obey this law, the energy of the substances inside the container must be equal to the substance added to it. The energy is in the form of heat. There can be two types of heat energy: latent heat and sensible heat. Sensible heat is energy added or removed when a substance changes in temperature. Latent heat is the energy added or removed at a constant temperature during a phase change. Since there is no mention of phase change, we assume the heat involved here is sensible heat. The equation for sensible heat is:

H = mCpΔT
where
m is the mass of the substance
Cp is the specific heat of a certain type of material or substance
ΔT is the change in temperature.

So the law of conservation of heat tells that:

Sensible heat of Z + Sensible heat of container = Sensible heat of X

Since we have no idea what these substances are, there is no way of knowing the Cp. We can't proceed with the calculations. So, we can only assume that in the duration of 15 minutes, the whole system achieves equilibrium. Therefore, the equilibrium temperature of the system is equal to 32°C. The answer is C.

5 0

2 years ago

A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea

Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
$R_T = R_1 + R_2 + ... + R_n$

In parallel:
$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

$\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega$

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
$R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}$

We can use Ohm's law to solve for the current in the circuit.

$i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}$

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

$V = iR\\\\V = (6)(15) = \boxed{90 V}$