an isolated container holding 5 kg of sulfuric acid is at 60 C. if 2.49x10^5 Joules of energy were transferred out of the acid, what would be the new temp. of the acid?
To determine the new temperature of the acid, we can use the equation:
\(Q = mcΔT\)
Where:
Q is the energy transferred (in Joules),
m is the mass of the substance (in kg),
c is the specific heat capacity of the substance (in J/kg·°C),
and ΔT is the change in temperature (in °C).
In this case:
Q = 2.49x10^5 J
m = 5 kg
The specific heat capacity of sulfuric acid is 1.34 J/g·°C, which can be converted to J/kg·°C by dividing by 1000.
c = 1.34 J/g·°C ÷ 1000 = 0.00134 J/kg·°C
Now we can rearrange the formula to solve for ΔT:
ΔT = Q / (mc)
Substituting the given values:
ΔT = 2.49x10^5 J / (5 kg × 0.00134 J/kg·°C)
ΔT = 2.49x10^5 J / 0.0067 J/°C
Thus, the change in temperature is ΔT ≈ 3.71x10^4 °C.
Finally, to find the new temperature, we add the change in temperature to the initial temperature of 60°C:
New temperature = 60°C + 3.71x10^4°C
Hence, the new temperature of the acid would be approximately 37,100°C.