a 1.9 kg block of iron at 26 degrees Celcius is rapidly heated by a torch such that 15 kJ is transferred to it.

Question

What temperature would the block of iron reach (assuming the complete transfer of heat and no loss to the surroundings)? If that same amount of heat (15 kJ ) was quickly transferred to a 850 g pellet of copper at 23 degrees celcius, what temperature would the copper reach before it begins losing heat to the surroundings? Use the equation for heat capacity and the following heat capacity values:

Cs, Fe(s) = 0.450 J / (g * C)

Cs, Cu(s) = 0.385 J / (g * C)

Answer 1

We will be using the specific heat capacity to answer this question. As detailed above the Q represents the heat energy usually in Joules or KJ, m represents the mass of the material usually having grams or kg as its units, C represents the specific heat capacity of the material you are solving for units Joules/(g*C), and delta T represents the change of temperature in Celsius or Kelvin units.

Solve:

Because we are working with J/g*C for our specific heat ratios we will need to convert our given information into those units.

Listing the information in appropriate units:

Mass of Fe- 1900 g

T initial- 26 C

Q- 15000 J

Solving for the T final plug in your corrected values:

Q=mc(T final- T initial)

15000J= 1900g(0.450J/g*C)(x-26C)

Solve for x

1)15000J= 855J/C (x- 26C)

I multiplied the grams by the specific heat ratio in the step above

2)15000J= 855J/C(x) – 22230J

I distributed the 855 J/C to the parenthesis

3) 15000J+22230J= 855 J/C(x)

37230 J= 855 J/C(x)

I added the Joules to the other side

4) 37230 J / 855 J/C=x

X= 43.54 Celsius

Divide 37230 J by 855 J/C, The Joules cancel and you are left with 43.54 Celsius as the final temperature of the iron block.

** I want you to try the second step of this problem as you use the same method to solve for T final when heating 850 g of Cu.

Answer 2

For iron:

q = mC∆T

q = heat =15 kJ = 15,000 J

m = mass = 1.9 kg = 1900 g

C = specific heat = 0.450 J/g/deg

∆T = change in temperature = ?

∆T = q/mC = 15,000 J / (1900 g)(0.450 J/g/deg)

∆T = 17.54 degrees = 18 degrees C

New temperature = 26 degrees + 18 degrees = 44 degrees C

For copper:

q = mC∆T

∆T = q/mC = 15,000 J/(850 g)(0.385 J/g/deg)

∆T = 45.8 degrees C = 46 degrees C

New temperature = 23 degrees + 46 degrees = 69 degrees C