A 22.44g sample of iron absorbs 180.8J of heat, upon which the tempature of the sample increases from 21.1 C to 39.0 C. What is the specific heat of iron?
Q=mc(t2-t1)
180.8=22.44 x c(39.0-21.1)
c= 0.450 Jg-1.C-1
q = mass x specific heat x (Tfinal-Tinitial).
Solve for specific heat.
To find the specific heat of iron, we can use the formula:
Heat (Q) = mass (m) * specific heat (C) * change in temperature (ΔT)
Given:
Mass (m) = 22.44g
Heat (Q) = 180.8J
Initial temperature (T₁) = 21.1°C
Final temperature (T₂) = 39.0°C
First, we need to convert the temperatures from Celsius to Kelvin:
T₁ = 21.1°C + 273.15 = 294.25K
T₂ = 39.0°C + 273.15 = 312.15K
Next, we can calculate the change in temperature (ΔT):
ΔT = T₂ - T₁ = 312.15K - 294.25K = 17.9K
Now, we can rearrange the formula to solve for the specific heat (C):
C = Q / (m * ΔT)
Substituting the given values:
C = 180.8J / (22.44g * 17.9K)
Performing the calculation:
C ≈ 0.44 J/g·K
Therefore, the specific heat of iron is approximately 0.44 J/g·K.
To find the specific heat of iron, we can use the formula:
q = m * c * ΔT
where:
q = heat absorbed (in joules)
m = mass of the sample (in grams)
c = specific heat of the substance (in J/g°C)
ΔT = change in temperature (in °C)
In the given problem, the heat absorbed (q) is 180.8 J, the mass of the iron sample (m) is 22.44 g, the initial temperature (T1) is 21.1°C, and the final temperature (T2) is 39.0°C.
First, we calculate the change in temperature (ΔT) by subtracting the initial temperature from the final temperature:
ΔT = T2 - T1
ΔT = 39.0°C - 21.1°C
ΔT = 17.9°C
Now, we can rearrange the formula to solve for the specific heat (c):
c = q / (m * ΔT)
Plugging in the values, we have:
c = 180.8 J / (22.44 g * 17.9°C)
Now we can calculate the specific heat:
c = 180.8 / (22.44 * 17.9)
c ≈ 0.44 J/g°C
Therefore, the specific heat of iron is approximately 0.44 J/g°C.