The radiator in your car uses either water or ethylene glycol (anti-freeze) to remove heat from the engine and dissipate it to the outside air. If the water recirculation rate is 17.0 kg per minute of water and the water temperature leaving the engine and entering the radiator is 86.0 C and the temperature of the water leaving the radiator and entering the engine is 46.0 C, then how much energy (kJ) is dissipated by the radiator per minute? The specific heat capacity of water is 4.18 J/g ·K. The units are kJ/minute. The error interval is +/- 5%.
q = mass x specific heat H2O x (Tfinal-Tinitial)
I would use mass = 17,000 and calculate q in J, then convert to kJ.
To calculate the amount of energy dissipated by the radiator per minute, we can use the formula:
Energy (kJ) = mass flow rate (kg/min) * specific heat capacity (kJ/kg·K) * temperature difference (K)
First, let's convert the given temperatures from degrees Celsius to Kelvin:
Temperature leaving the engine and entering the radiator (C) = 86.0 °C
Temperature leaving the engine and entering the radiator (K) = 86.0 °C + 273.15 = 359.15 K
Temperature leaving the radiator and entering the engine (C) = 46.0 °C
Temperature leaving the radiator and entering the engine (K) = 46.0 °C + 273.15 = 319.15 K
Now, let's calculate the temperature difference:
Temperature difference (ΔT) = Temperature leaving the engine - Temperature leaving the radiator
ΔT = 359.15 K - 319.15 K = 40 K
Next, let's calculate the amount of energy dissipated:
Energy (kJ) = mass flow rate * specific heat capacity * temperature difference
Given:
Mass flow rate = 17.0 kg/min
Specific heat capacity of water = 4.18 J/g·K
To convert the specific heat capacity from J/g·K to kJ/kg·K:
Specific heat capacity (kJ/kg·K) = 4.18 J/g·K * (1 kJ/1000 J) * (1 kg/1000 g) = 0.00418 kJ/kg·K
Now, let's substitute the values into the formula:
Energy (kJ) = 17.0 kg/min * 0.00418 kJ/kg·K * 40 K = 2.8264 kJ/min
Therefore, the amount of energy dissipated by the radiator per minute is approximately 2.8264 kJ/min.
Note: The error interval is given as +/- 5%, so let's calculate the maximum and minimum values:
Maximum energy dissipated = 2.8264 kJ/min + 5% = 2.8264 kJ/min + 0.1413 kJ/min = 2.9677 kJ/min
Minimum energy dissipated = 2.8264 kJ/min - 5% = 2.8264 kJ/min - 0.1413 kJ/min = 2.6851 kJ/min
Therefore, the actual amount of energy dissipated by the radiator per minute falls within the range of approximately 2.6851 kJ/min to 2.9677 kJ/min.