A 10.0 grams piece of gold 90.0 Celsius was stirred into 15.0ml of water at an initial temperature of 25.0 degrees Celsius. The final temperature of the water was 26.34 degrees celsius. Calculate the specific heat of gold. The specific heat of water is 4.184 J/g- degrees celsius
recall that the amount of heat absorbed or released is given by:
Q = mc(T2-T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-K)
T = temperature (in C or K)
*note: Q = (+) when heat is absorbed and (-) when heat is released
to get the final temp, we note that Q,released = Q,absorbed. therefore:
Q,released = Q,absorbed
-m1*c1*(T2-T,r1) = m2*c2*(T2-T,a1)
we're solving for c1, which is the specific heat capacity of gold.
assuming that the density of water is 1g/mL, the mass of water (for a volume equal to 15 mL) is equal to 15 g. substituting,
-10*c1*(26.34-90) = 15*4.184*(26.34-25)
636.6*c1 = 84.0984
c1 = 0.1321 J/g-K
hope this helps~ :)
To calculate the specific heat of gold, we can use the formula:
q = mcΔT
Where:
q = heat energy transferred
m = mass of the substance
c = specific heat capacity
ΔT = change in temperature
In this case, we know the mass of the gold is 10.0 grams, and the change in temperature (ΔT) for the water is (26.34 degrees Celsius - 25.0 degrees Celsius) = 1.34 degrees Celsius.
Now, we need to calculate the amount of heat energy transferred (q) to the water. We can use the formula:
q = mcΔT
Given that the mass of the water is 15.0 grams and the specific heat capacity of water (c) is 4.184 J/g- degrees Celsius, we can substitute these values into the formula:
q = (15.0 grams) × (4.184 J/g- degrees Celsius) × (1.34 degrees Celsius)
q = 89.028 J
Next, we can calculate the specific heat of gold (c) by rearranging the formula:
c = q / (m × ΔT)
Substituting the known values:
c = (89.028 J) / (10.0 grams × 1.34 degrees Celsius)
c ≈ 6.64 J/g- degrees Celsius
Therefore, the specific heat of gold is approximately 6.64 J/g- degrees Celsius.