7750 Joules of energy is added to a 325 grams stone at 24.0 degrees centigrade. The temperature increases by 75.0 degrees centigrade. What is the specific heat of the stone?
q = mass stone x specific heat x delta T
Solve for specific heat.
To find the specific heat of the stone, we need to use the equation:
Q = m * c * ΔT
where:
Q = energy added (in Joules)
m = mass of the stone (in grams)
c = specific heat of the stone (in J/g°C)
ΔT = change in temperature (in °C)
Given:
Q = 7750 J
m = 325 g
ΔT = 75.0 °C
Rearranging the equation, we can solve for c:
c = Q / (m * ΔT)
Substituting the given values:
c = 7750 J / (325 g * 75.0 °C)
Now, let's calculate the specific heat:
c = 0.304 J/g°C
Therefore, the specific heat of the stone is approximately 0.304 J/g°C.