1) Calculate the specific heat capacity (J/g oC) of iron, if a 803 g sample lost 937.4 J of heat energy in going from an initial temperature of 5.50 oC to a final temperature of 2.90 oC.
2)Calculate the final temperature (oC) of a 243 g sample of water originally at a temperature of 84.6 oC that lost 73640 J of heat energy.
1.
-q = mass x sp.h. x (Tfinal-Tinitial)
2.
-q = mass x sp.h. x (Tfinal-Tinitial)
Calculate the final temperature of 290mL of water initially at 31∘C upon absorption of 20kJ of heat.
To calculate the specific heat capacity of iron in question 1, we can use the equation:
q = m * c * ΔT
where:
- q is the heat energy (in joules),
- m is the mass of the sample (in grams),
- c is the specific heat capacity (in J/g oC), and
- ΔT is the change in temperature (final temperature - initial temperature).
Given:
- mass (m) = 803 g
- initial temperature = 5.50 oC
- final temperature = 2.90 oC
- heat energy (q) = -937.4 J (negative sign indicates heat loss)
Substituting the given values into the equation, we can solve for the specific heat capacity (c):
-937.4 J = 803 g * c * (2.90 oC - 5.50 oC)
First, calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 2.90 oC - 5.50 oC
ΔT = -2.60 oC (negative indicates a decrease in temperature)
Now, rearrange the equation to solve for c:
c = -937.4 J / (803 g * -2.60 oC)
c ≈ 0.177 J/g oC (rounded to three significant figures)
Therefore, the specific heat capacity of iron is approximately 0.177 J/g oC.
Now, for question 2, to calculate the final temperature, we can again use the equation:
q = m * c * ΔT
Given:
- mass (m) = 243 g
- initial temperature = 84.6 oC
- heat energy (q) = -73640 J (negative sign indicates heat loss)
We want to find the final temperature (ΔT). Rearranging the equation, we have:
ΔT = q / (m * c)
Substituting the given values into the equation:
ΔT = -73640 J / (243 g * c)
Here, we'll use the specific heat capacity of water, which is approximately 4.18 J/g oC.
ΔT = -73640 J / (243 g * 4.18 J/g oC)
ΔT ≈ -73.94 oC (rounded to two decimal places)
Finally, calculate the final temperature:
final temperature = initial temperature + ΔT
final temperature ≈ 84.6 oC + (-73.94 oC)
final temperature ≈ 10.66 oC (rounded to two decimal places)
Therefore, the final temperature of the water sample is approximately 10.66 oC.