Calorimetry Homework: Rate of Heat Gain Before Switch Off

[umm...]

Homework Statement

when a metal cylinder of mass 0.02kg and specific heat capacity 500J/kg/K is heated by an electric heater working at a constant power, the initial rate of rise of temperature is 3K/min. After a time the heater is switched off and the initial rate of fall of temperature is 0.3K/min. what is the rate at which the cylinder gains heat energy immediately before the heater is switched off? ans 0.45w

Homework Equations

Q=mc(dT)

The Attempt at a Solution

heat gained per second
Q=0.02x500x3/60=0.5W
Q=0.02x500x0.3/60=0.05W
0.5-0.05=0.45W
i did this just tried but didnt get the concept of the questions...and not even sure if the method is correct!
could someone please elaborate the qn to me!

 

[anathema]

Hello, thank you for your question. Let's break down the question and go through the steps together.

First, we are given a metal cylinder with a mass of 0.02kg and a specific heat capacity of 500J/kg/K. This means that it takes 500 Joules of heat energy to raise the temperature of 1kg of this metal by 1 Kelvin.

Next, we are told that the cylinder is being heated by an electric heater at a constant power. This means that the rate at which heat energy is being supplied to the cylinder is constant.

We are also given the initial rate of rise of temperature, which is 3K/min. This means that for every minute, the temperature of the cylinder is increasing by 3 Kelvin.

Now, here comes the tricky part. The heater is switched off after a certain amount of time, and we are given the initial rate of fall of temperature, which is 0.3K/min. This means that after the heater is switched off, the temperature of the cylinder is decreasing by 0.3 Kelvin every minute.

The question is asking for the rate at which the cylinder gains heat energy immediately before the heater is switched off. This means we need to find the rate of heat energy being supplied to the cylinder at that specific moment.

To do this, we can use the equation Q=mc(dT), where Q is the heat energy, m is the mass of the cylinder, c is the specific heat capacity, and (dT) is the change in temperature.

Using the information given, we can calculate the heat energy gained per second before the heater is switched off. This is done by dividing the heat energy gained per minute by 60 seconds. So, for the initial rate of rise of temperature, we have:

Q = (0.02kg)(500J/kg/K)(3K/min)/60s = 0.5W

And for the initial rate of fall of temperature, we have:

Q = (0.02kg)(500J/kg/K)(0.3K/min)/60s = 0.05W

Now, to find the rate at which the cylinder gains heat energy immediately before the heater is switched off, we need to subtract the rate of heat energy being lost (0.05W) from the rate of heat energy being gained (0.5W). This gives us:

0.5W - 0.05W = 0.