Heat water; mechanical (the movement of a turbine is based off of mechanical energy, not chemical or potential).
Answer:
P = 164 Atm
Explanation:
PV = nRT => P = nRT/V
n = 10.0 moles
R = 0.08206 L·Atm/mol·K
T = 27.0°C = 300 K
V = 1.50 Liters
P = (10.0 mol)(0.08206 L·Atm/mol·K )(300 K)/(1.50 Liters) = 164.12 Atm ≅ 164 Atm (3 sig. figs.)
Answer is: a) I only.
Above critical temperature of CO₂, a gas cannot be liquefied no matter how much pressure is applied. Temperature and pressure above its critical point is called supercritical fluid and this is <span>intermediate between gaseous and liquid states.</span>
Answer: n∗R=22+273.15/4.2∗5n
P2=n∗R∗T2/V2=n∗R∗33.6+273.15/10
Explanation:
Answer:
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Explanation:
Step 1: Data given
Initial temperature = 10.0 °C
Final temperature = 25.0 °C
Energy required = 30000 J
Mass of the object = 40.0 grams
Step 2: Calculate the specific heat capacity of the object
Q = m* c * ΔT
⇒With Q = the heat required = 30000 J
⇒with m = the mass of the object = 40.0 grams
⇒with c = the specific heat capacity of the object = TO BE DETERMINED
⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C
30000 J = 40.0 g * c * 15.0 °C
c = 30000 J / (40.0 g * 15.0 °C)
c = 50 J/g°C
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
