A metal rod is made of 26kg of iron, how much heat must be supplied to the radiator to raise its temperature from 25 Celsius to 55 Celsius?

My answer is
Q= 350, 220J

Follow up question
If a boiler generateswater at 79.5 Celsius what mass of water is needed to provide the heat required in the previous problem?

On the first, depends on whay you used as specfic heat capacity. I used .46kJ/kgC, and got 358kJ

I dont understand on the second the question, as what is the initial temp of the water. It makes no sense to me.

The first and second question are related sir. But i cant see the use of the initial temp on the second question. To solve for the mass on the 2nd question what change in temp will i use?

I agree, you need water initial temp. I am also confused calling it a boiler generating water. Boilers generate steam.

Youre right sir. Am i going to use this eqn

Q=mc deltaT
Where my change in temp will be 79.5-55 C?

Again, maybe, I have no idea what it meant.

Okay thank you sir

To determine the heat required to raise the temperature of the metal rod, you can use the formula:

Q = mcΔT

Where:
Q is the heat energy
m is the mass of the object
c is the specific heat capacity of the material
ΔT is the change in temperature

In this case, the metal rod is made of iron. The specific heat capacity of iron is approximately 450 J/kg°C.

Given:
m = 26 kg (mass of the iron rod)
ΔT = (55°C - 25°C) = 30°C (change in temperature)

Using the formula, plug in the values to calculate the heat energy:

Q = 26 kg * 450 J/kg°C * 30°C
Q = 351,000 J

So, the heat required to raise the temperature of the metal rod from 25°C to 55°C is 351,000 J (or 351 kJ, kilojoules).

Now, let's move on to the follow-up question.

To determine the mass of water needed to provide the same amount of heat energy, we rearrange the formula:

Q = mcΔT

Since the heat energy (Q) is the same as before (351,000 J), and we know the specific heat capacity of water is approximately 4,186 J/kg°C, we can rearrange the formula to solve for mass (m):

m = Q / (cΔT)

m = 351,000 J / (4,186 J/kg°C * 30°C)

m ≈ 2.65 kg

So, approximately 2.65 kg of water is needed to provide the heat required in the previous problem.