2.00 grams of hydrocarbon warmed 150.0 ml. water from 25.0 degree C to 30.0 degree C. Calculate the heat of hydrocarbon in K/g. the specific heat of water is 4.184 J/g. degree C
Did you intend to type kJ/g instead of K/g? And I assume the hydrocarbon warmed the water because it was burned.
q heat released = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q/2.00 = J/gram. Convert to kJ/g.
To solve this problem, we will use the formula for heat transfer:
Q = mcΔT
where:
Q is the heat transferred,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
ΔT is the change in temperature.
First, let's calculate the heat transferred to the water:
Q(water) = mcΔT
= (150.0 g)(4.184 J/g°C)(30.0°C - 25.0°C)
= 6,276 J
Next, using the formula Q = mcΔT, we can solve for the heat of the hydrocarbon:
Q(hydrocarbon) = mcΔT
= (2.00 g)(c)(30.0°C - 25.0°C)
The specific heat capacity (c) for the hydrocarbon is unknown. To find it, we can rearrange the formula and solve for c:
c = Q(hydrocarbon) / (mΔT)
= 6,276 J / (2.00 g)(30.0°C - 25.0°C)
Now, substitute the values into the equation to find the specific heat capacity of the hydrocarbon:
c = 6,276 J / (2.00 g)(5.0°C)
= 6,276 J / 10.00 g°C
= 627.6 J/g°C
Therefore, the heat of the hydrocarbon is 627.6 J/g°C.