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?
show steps please!
To determine the new temperature of the sulfuric acid after transferring 2.49x10^5 Joules of energy out of it, we can use the formula for heat transfer:
Q = mcΔT
Where:
Q is the heat energy transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, we need to calculate the amount of heat energy transferred:
Q = 2.49x10^5 J
Next, we need to find the specific heat capacity of sulfuric acid. The specific heat capacity of sulfuric acid is approximately 1.38 J/g·°C.
Now, let's substitute the known values into the formula and solve for ΔT:
2.49x10^5 J = (5 kg) * (1.38 J/g·°C) * ΔT
Let's simplify the equation:
2.49x10^5 J = 5 kg * (1380 J/kg·°C) * ΔT
2.49x10^5 J = 6900 J/°C * ΔT
Divide both sides of the equation by 6900 J/°C:
2.49x10^5 J / 6900 J/°C = ΔT
ΔT ≈ 36.09 °C
Therefore, the new temperature of the sulfuric acid would be approximately 60 °C - 36.09 °C, which is approximately 23.91 °C.