What is the specific heat of a substance if 26.0 kJ are required to raise the temperature of 250. g of the substance from 34.0°C to 201.0°C?
q = mass x specific heat x delta T
To find the specific heat of a substance, we can use the following formula:
Q = m * c * ΔT
Where:
Q = amount of heat transferred
m = mass of the substance
c = specific heat of the substance
ΔT = change in temperature
In this case, we are given:
Q = 26.0 kJ (kilojoules)
m = 250 g (grams)
ΔT = (201.0°C - 34.0°C) = 167.0°C
Now, let's substitute the values into the formula and solve for c:
26.0 kJ = 250 g * c * 167.0°C
To solve for c, we need to rearrange the formula to isolate c:
c = Q / (m * ΔT)
Let's plug in the given values:
c = 26.0 kJ / (250 g * 167.0°C)
First, we need to convert kJ to J (joules) since the units of c are J/(g°C). Since 1 kJ = 1000 J, we have:
c = (26.0 kJ * 1000 J/kJ) / (250 g * 167.0°C)
c = 26000 J / (250 g * 167.0°C)
Now, let's calculate the specific heat:
c = 0.988 J/(g°C)
Therefore, the specific heat of the substance is 0.988 J/(g°C).