What is the specific heat of a substance if 35.5 kJ are required to raise the temperature of 345 g of the substance from 69.0°C to 150.0°C?
All of these can be done with one equation.
q = mass x specific heat x (Tfinal-Tinitial).
To calculate the specific heat of a substance, you can use the formula:
q = m * c * ΔT
Where:
q is the heat energy absorbed or released
m is the mass of the substance
c is the specific heat of the substance
ΔT is the change in temperature
In this case, the heat energy (q) is given as 35.5 kJ (kilojoules). The mass (m) is 345 g. The change in temperature (ΔT) can be calculated by subtracting the initial temperature (69.0°C) from the final temperature (150.0°C).
First, convert the given heat energy to joules:
1 kJ = 1000 J
So, 35.5 kJ = 35.5 * 1000 J = 35,500 J
Next, calculate ΔT:
ΔT = final temperature - initial temperature
ΔT = 150.0°C - 69.0°C = 81.0°C
Now that we have all the values, we can rearrange the formula to solve for c:
c = q / (m * ΔT)
Substituting the values:
c = 35,500 J / (345 g * 81.0°C)
To proceed with the calculation, we need to convert the mass to kilograms and the temperature difference to Kelvin.
Converting mass:
1 g = 0.001 kg
345 g = 0.345 kg
Converting temperature difference:
ΔT(K) = ΔT(°C) + 273.15
ΔT(K) = 81.0°C + 273.15 K = 354.15 K
Now, substitute the converted values into the formula:
c = 35,500 J / (0.345 kg * 354.15 K)
Calculate the specific heat:
c = 35,500 J / 121.95075 K
Round the result to an appropriate number of significant figures (if necessary) to get the final answer for the specific heat of the substance.