Answer:
V₂ = 45.53 L
Explanation:
Given data:
Initial temperature = 850 K
Initial volume = 65 L
Initial pressure = 450 KPa
Final temperature = 430 K
Final pressure = 325 KPa
Final volume = ?
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 450 KPa× 65 L × 430 K / 850 K × 325KPa
V₂ = 12577500 KPa .L. K / 276250 K. KPa
V₂ = 45.53 L
Answer:
i think the answer is letter C. From 35°c to 45°c
Explanation:
sorry if it is wrong
It shows the atomic number
Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
