The specific heat capacity of gold at 25°C is 1.290×10-1 J/g/K.
For a 5.25×102 g sample of gold, how much will the temperature increase if 6.908×102 J of energy is put into the system?
(Assume that the heat capacity is constant over this temperature range.)
q = mass Au x specific heat Au x (Tfinal-Tinitial)
690.8 = 525 x 0.1290 x delta T.
Solve for delta T.
To calculate the temperature increase, you can use the formula:
ΔT = Q / (m * C)
Where:
ΔT = Temperature increase (in °C or K)
Q = Energy added to the system (in J)
m = Mass of the sample (in g)
C = Specific heat capacity of the substance (in J/g/K)
Given values:
Q = 6.908×10^2 J
m = 5.25×10^2 g
C = 1.290×10^-1 J/g/K
Plugging in the values into the formula, we get:
ΔT = (6.908×10^2 J) / (5.25×10^2 g * 1.290×10^-1 J/g/K)
Now, let's simplify the equation:
ΔT = (6.908×10^2 J) / (6.7825×10^1 J/K)
Finally, we can calculate ΔT:
ΔT ≈ 10.18 K
Therefore, the temperature will increase by approximately 10.18 degrees Celsius (or Kelvin) if 6.908×10^2 J of energy is put into the system.
To determine the increase in temperature, we can use the specific heat capacity formula:
q = mcΔT
Where:
q = heat energy (in Joules)
m = mass (in grams)
c = specific heat capacity (in J/g/K)
ΔT = change in temperature (in Kelvin)
First, we need to convert the given mass from grams to kilograms to match the unit of specific heat capacity (J/g/K). We can do this by dividing the mass by 1000:
mass = 5.25×10^2 g = 5.25×10^2 g / 1000 g/kg = 5.25 kg
Now, we can rearrange the formula to solve for ΔT:
ΔT = q / (mc)
Substituting the given values:
q = 6.908×10^2 J
m = 5.25 kg
c = 1.290×10^-1 J/g/K = 1.290×10^-1 J/g/K * 1000 g/kg = 1.290×10^2 J/kg/K
ΔT = (6.908×10^2 J) / ((5.25 kg)(1.290×10^2 J/kg/K))
Calculating ΔT:
ΔT = 1.0303 K
Therefore, the temperature will increase by approximately 1.0303 Kelvin.