Heating Water Using Circuit Resistor

[Canadian]

Homework Statement

In a circuit a 20-ohm resistor sits inside 104 g of pure water that is surrounded by insulating Styrofoam, there are additional resistors in the circuit.

If the water is initially at temperature 10.5 degrees celcius, how long will it take for its temperature to rise to 58.5 degrees celcius?
Use 4190 as the heat capacity of water

Homework Equations

V=IR P=VI

The Attempt at a Solution

I assumed that the styrofoam was a "perfect" insulator did not absorb any heat or allow any to escape.

Does this work.

 

[Andrew Mason]

Homework Equations

V=IR P=VI

The Attempt at a Solution

I assumed that the styrofoam was a "perfect" insulator did not absorb any heat or allow any to escape.

Does this work.

Your work and the answer are fine, except that you have too many significant figures in your result.

AM

 

[m2003]

I cannot provide a definitive answer without more information about the circuit and its components. However, I can offer some thoughts and suggestions for finding a solution.

First, it is important to note that the heat capacity of water (4190 J/kg*K) is a crucial piece of information in this problem. This value tells us how much energy is required to raise the temperature of 1 kilogram of water by 1 degree Kelvin. In this case, we are dealing with 104 grams of water, so we need to adjust the heat capacity accordingly.

Next, we need to consider the power dissipated by the resistors in the circuit. The power equation (P=VI) tells us that the power dissipated by a resistor is equal to the voltage across it multiplied by the current flowing through it. In this problem, we know the resistance of the 20-ohm resistor, but we need to know the voltage and current to calculate the power. This information may be provided in the problem or can be calculated using Ohm's Law (V=IR).

Once we have the power dissipated by the resistor, we can use the formula Q=mcΔT to calculate the energy (Q) required to raise the temperature of the water by a certain amount (ΔT). In this case, we know the mass of water (104 grams) and the specific heat capacity (4190 J/kg*K), but we need to calculate the change in temperature (ΔT) from 10.5 degrees Celsius to 58.5 degrees Celsius.

Finally, we can use the relationship between energy and power (P=Q/t) to find the time (t) required for the water to reach the desired temperature. This formula tells us that the power dissipated by the resistor over a certain amount of time (t) is equal to the energy (Q) required to raise the temperature of the water.

In summary, to solve this problem, we need to know the voltage and current in the circuit, the power dissipated by the resistors, and the change in temperature of the water. With this information, we can use the equations mentioned above to calculate the time required for the water to reach 58.5 degrees Celsius. It is also important to consider any other factors that may affect the heating of the water, such as heat loss to the surroundings.