when 435J of heat is added to 3.40grams of olive oil at 21c the temperature increases 85.0c
I don't see a question but I assum you want to solve for the specific heat of the olive oil.
q = mc*deltaT
435 = 3.40 g * specific heat x 85 C
Solve for specific heat in J/g*C
Post your work if you get stuck.
To solve this problem, we need to use the formula for heat transfer:
Q = mcΔT
Q represents the amount of heat transferred, m represents the mass of the substance, c represents the specific heat capacity of the substance, and ΔT represents the change in temperature.
First, let's find the specific heat capacity of olive oil. The specific heat capacity of a substance is the amount of heat energy required to change the temperature of 1 gram of that substance by 1 degree Celsius.
The specific heat capacity of olive oil is approximately 2.0 J/(g·°C).
Now, let's calculate the heat transferred using the formula:
Q = mcΔT
Q = (3.40 g) * (2.0 J/g·°C) * (85.0 °C - 21 °C)
Q = 3.40 g * 64 °C * 2.0 J/g·°C
Q = 435.2 J
The amount of heat transferred is 435.2 Joules (J).
Therefore, when 435 J of heat is added to 3.40 grams of olive oil at 21°C, the temperature increases by 85.0°C.
To calculate the specific heat capacity of olive oil, you can use the equation:
q = m * c * ΔT
Where:
q = heat energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g·°C)
ΔT = change in temperature (in °C)
Given:
q = 435 J
m = 3.40 g
ΔT = 85.0 °C
Rearranging the equation, we can solve for c:
c = q / (m * ΔT)
Substituting the given values:
c = 435 J / (3.40 g * 85.0 °C)
Now, let's calculate:
c = 435 J / (289 g·°C)
c ≈ 1.51 J/g·°C
Therefore, the specific heat capacity of olive oil is approximately 1.51 J/g·°C.