given 10.8 grams of substance y,if the substance absorbs 231 kilojoules of energy and the specfic heat of substance is 6.55 J/g*0C what is the heat change
q = mass x specific heat x delta T
q = 231
mass = 10.8
sp.h. = 6.55
dT= solve
3.265 deg. C
To find the heat change of substance Y, we can use the formula:
q = m * c * ΔT
Where:
q = Heat change (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)
In this case, we are given:
m = 10.8 grams
c = 6.55 J/g°C
ΔT is not given, so we'll assume there is a change in temperature.
However, we are also given that the substance absorbs 231 kilojoules of energy. To convert kilojoules to joules, we need to multiply by 1000:
231 kilojoules * 1000 = 231,000 joules
Now, we can rearrange the formula to find ΔT:
q = m * c * ΔT
231,000 joules = 10.8 grams * 6.55 J/g°C * ΔT
Simplifying the equation:
231,000 joules = 70.74 grams*J/°C * ΔT
Now, we can solve for ΔT:
ΔT = 231,000 joules / (10.8 grams * 6.55 J/g°C)
≈ 3347.22 °C
Therefore, the heat change of substance Y is approximately 3347.22 °C.