Question

410A is a commonly used refrigerant. What

is the energy change when 100. g of 410A
goes from -40.0 °C to -60.0 °C?
Heating Curve Data for Coolant 410a
Boiling Point (°C)
-51.53
AHVan (kJ/kg)
276.2
vap
Specific Heat, gas (kJ/kg K)
0.823
Specific Heat, liquid (kJ/kg K)
1.84
q = [?] kJ
Do not round until the end. Include either a + or -
sign AND the magnitude.

Answer 1

Answer: The energy change when 100 g of 410A goes from -40.0 °C to -60.0 °C is -184 kJ.

Step-by-step explanation:

To find the energy change when 100 g of 410A goes from -40.0 °C to -60.0 °C, we need to use the specific heat capacity of 410A in the liquid state, which is 1.84 kJ/kg K.

The energy change can be calculated using the formula

q = m * c * ΔT,

where q is the energy change, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

Plugging in the values, we get:

q = 100 g * 1.84 kJ/kg K * (-60.0 °C - (-40.0 °C))

Simplifying the equation, we get:

q = 100 g * 1.84 kJ/kg K * (-100.0 °C)

Converting 100 g to kilograms gives us:

q = 0.1 kg * 1.84 kJ/kg K * (-100.0 °C)

Solving for q, we get:

q = -184 kJ

So the energy change when 100 g of 410A goes from -40.0 °C to -60.0 °C is -184 kJ.