A 4.90 g nugget of pure gold absorbed 255 J of heat. What was the final temperature of the gold if the initial temperature was 24.0°C? The specific heat of gold is 0.129 J/(g·°C).
q = mass Au x specific heat Au x (Tfinal-Tinitial)
463.5342
YOU CAN USE ALGEBRA!
energy absorbed=given grams x specific heat x (unknown final temperature-initial temp)
SO:
255=4.90(.129)(x-24)
255=0.6321(x-24)
DIVIDE BOTH SIDES BY .6321
.6321(255)=(.6321(x-24))/,6321
403.42 = x-24
+24 +24
----------------
x=427.42
To find the final temperature of the gold, we can use the formula for heat transfer:
q = m * c * ΔT
where:
- q is the heat transfer (in joules)
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in J/(g·°C))
- ΔT is the change in temperature (in °C)
In this case, we are given:
- q = 255 J
- m = 4.90 g
- c = 0.129 J/(g·°C)
- ΔT = final temperature - initial temperature
First, let's rearrange the formula to solve for ΔT:
ΔT = q / (m * c)
Substituting the given values:
ΔT = 255 J / (4.90 g * 0.129 J/(g·°C))
Now, let's calculate ΔT:
ΔT = 255 J / (4.90 * 0.129 (J/(g·°C)))
ΔT ≈ 396.90 °C (rounded to 2 decimal places)
Finally, to find the final temperature, we need to add ΔT to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 24.0 °C + 396.90 °C
Therefore, the final temperature of the gold is approximately 420.90 °C (rounded to 2 decimal places).