Answer:
a) 5.63 atm
Explanation:
We can use combined gas law
<em>The combined gas law</em> combines the three gas laws:
- Boyle's Law, (P₁V₁ =P₂V₂)
- Charles' Law (V₁/T₁ =V₂/T₂)
- Gay-Lussac's Law. (P₁/T₁ =P₂/T₂)
It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.
P₁V₁/T₁ =P₂V₂/T₂
where P = Pressure, T = Absolute temperature, V = Volume occupied
The volume of the system remains constant,
So, P₁/T₁ =P₂/T₂
a) 
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
Answer:the resistance decrease
Explanation:
Answer:
+131Joules
Explanation:
Energy can be expressed using below expresion.
ΔE = (q + w).........eqn(1)
q will be + be if heat is gained hence, q= 240 J
work "w" will be - ve if work is done by the system, hence w= -109 J
Then substitute into eqn(1)
Change in Internal energy=
= (240 -109 )
= +131J
