You add 8950 J of heat to 174 g of iron. What is the temperature increase of the iron (c = 452 J/(kg • C°))?
To find the temperature increase of the iron, we can use the formula:
\(Q = mcΔT\)
Where:
- \(Q\) is the heat energy added (8950 J)
- \(m\) is the mass of the iron (174 g)
- \(c\) is the specific heat capacity of iron (452 J/(kg • C°))
- \(ΔT\) is the change in temperature (what we want to find)
First, let's convert the mass of the iron from grams to kilograms:
\(174 \, \text{g} = 0.174 \, \text{kg}\)
Now, we can rearrange the formula to solve for \(ΔT\):
\(ΔT = \frac{Q}{mc}\)
Substituting the values:
\(ΔT = \frac{8950 \, \text{J}}{0.174 \, \text{kg} \times 452 \, \text{J/(kg • C°)}}\)
Calculating:
\(ΔT \approx 28.82 \, \text{C°}\)
Therefore, the temperature of the iron increases by approximately 28.82 degrees Celsius.
To calculate the temperature increase of the iron, we need to use the specific heat capacity formula:
Q = m * c * ΔT
Where:
Q is the amount of heat energy (in joules) added to the substance,
m is the mass of the substance (in kilograms),
c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius), and
ΔT is the temperature change (in degree Celsius).
We are given:
Q = 8950 J (amount of heat energy added)
m = 174 g = 0.174 kg (mass of iron)
c = 452 J/(kg • C°) (specific heat capacity)
To find ΔT, rearrange the formula:
ΔT = Q / (m * c)
Substituting the given values into the formula:
ΔT = 8950 J / (0.174 kg * 452 J/(kg • C°))
Now, let's calculate it:
ΔT ≈ 8950 J / (0.174 kg * 452 J/(kg • C°))
ΔT ≈ 32.6711 C°
Therefore, the temperature of the iron will increase by approximately 32.67 degrees Celsius.
8950 J = .174 kg (delta T)[452 J/(kg deg C)]
so
delta T = 8950 /(.174*452)