Answer:
V₂= 1 L
Explanation:
Given that
Volume occupies V₁= 6 L
Initial pressure = P₁
Initial temperature = T₁
The final pressure =P₂ = 2 P₁
Final volume =V₂
Final temperature = T₁/3
As we know that equation for ideal gas
P V = m R T
P=pressure, V=volume, T=temperature
m=mass ,R=gas constant
Now from mass conservation
6 = 3 x 2 V₂
V₂= 1 L
So the final volume will be 1 L
Answer:
2274 J/kg ∙ K
Explanation:
The complete statement of the question is :
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
= mass of metal = 400 g
= specific heat of metal = ?
= initial temperature of metal = 100 °C
= mass of aluminum cup = 100 g
= specific heat of aluminum cup = 900.0 J/kg ∙ K
= initial temperature of aluminum cup = 15 °C
= mass of water = 500 g
= specific heat of water = 4186 J/kg ∙ K
= initial temperature of water = 15 °C
= Final equilibrium temperature = 40 °C
Using conservation of energy
heat lost by metal = heat gained by aluminum cup + heat gained by water