Answer:
209.3 Joules require to raise the temperature from 10 °C to 15 °C.
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m × c × ΔT
Given data:
mass of water = 10 g
initial temperature T1= 10 °C
final temperature T2= 15 °C
temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C
Energy or joules added to increase the temperature Q = ?
Solution:
We know that specific heat of water is 4.186 J/g .°C
Q = m × c × ΔT
Q = 10 g × 4.186 J/g .°C × 5 °C
Q = 209.3 J
Answer:
2726.85 °C
Explanation:
Given data:
Initial pressure = 565 torr
Initial temperature = 27°C
Final temperature = ?
Final pressure = 5650 torr
Solution:
Initial temperature = 27°C (27+273 = 300 K)
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
T₂ = P₂T₁ /P₁
T₂ = 5650 torr × 300 K / 565 torr
T₂ = 1695000 torr. K /565 torr
T₂
= 3000 K
Kelvin to degree Celsius:
3000 K - 273.15 = 2726.85 °C
Answer:
V2 = 894.4mL
Explanation:
P1= 124.1, V1= 578mL, P2 = 80.2kPa, V2= ?
Applying Boyle's law
P1V1 = P2V2
Substitute and simplify
124.1*578=80.2*V2
V2= 894.4mL
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