Because of their different longitude.
Answer:
–1647.45 J
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 112.5 g
Initial temperature (T₁) = 12.5 °C
Final temperature (T₂) = 9°C
Specific heat capacity (C) = 4.184 J/g°C
Heat (Q) absorbed =?
Next, we shall determine the change in temperature. This can be obtained as follow:
Initial temperature (T₁) = 12.5 °C
Final temperature (T₂) = 9°C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 9 – 12.5
ΔT = –3.5 °C
Finally, we shall determine the heat absorbed. This can be obtained as follow:
Mass (M) = 112.5 g
Change in temperature (ΔT) = –3.5 °C
Specific heat capacity (C) = 4.184 J/g°C
Heat (Q) absorbed =?
Q = MCΔT
Q = 112.5 × 4.184 × –3.5
Q = –1647.45 J
Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
The most common types of radiation are alpha particles, beta particles, and gamma rays
The first one because some people might not believe that and others might believe that.
