What is the specific heat of an unknown substance if a 2.45g samples releases 12.9 cal as its temperature changes from 25.1 degrees C to 20.1 degrees C?
q = mass x specific heat x (Tfinal-Tinitial)
To determine the specific heat of an unknown substance, we can use the formula:
q = mcΔT
Where:
q is the heat absorbed or released by the substance
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we are given:
mass (m) = 2.45 g
heat released (q) = -12.9 cal (negative because heat is released)
initial temperature (T1) = 25.1 °C
final temperature (T2) = 20.1 °C
First, we need to convert the heat from calories to joules because the specific heat unit is usually given in joules per gram per degree Celsius (J/g°C).
1 cal = 4.184 J (approximately)
So, the heat released in joules (q) is:
q = -12.9 cal × 4.184 J/cal ≈ -53.9016 J
Now, we can substitute the known values into the formula and solve for c:
-53.9016 J = (2.45 g) × c × (20.1°C - 25.1°C)
To simplify, subtract the temperatures:
-53.9016 J = (2.45 g) × c × (-5°C)
Next, divide both sides of the equation by -5°C:
-53.9016 J / -5°C = (2.45 g) × c
Now, divide both sides of the equation by (2.45 g):
(-53.9016 J / -5°C) / 2.45 g = c
Calculate the result:
c ≈ 4.419 J/g°C
Therefore, the specific heat of the unknown substance is approximately 4.419 J/g°C.