Answer:
The answer to your question is: V2 = 1.94 l
Explanation:
Data
V1 = 2.42 l
T1 = 25°C
P1 = 1 atm
V2 = ?
T2 = 25 -11 = 14°C
P2 = 1(0.7) = 0.7 atm
Formula
P1V1/T1 = P2V2/T2
Clear V2 from the equation
V2 = P1V1T2/ P2T1
V2 = (1)((2.42)(14) / (0.7)(25)
V2 = 33.88 / 17.5
V2 = 1.94 l
Answer:
c =0.2 J/g.°C
Explanation:
Given data:
Specific heat of material = ?
Mass of sample = 12 g
Heat absorbed = 48 J
Initial temperature = 20°C
Final temperature = 40°C
Solution:
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
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 40°C -20°C
ΔT = 20°C
48 J = 12 g×c×20°C
48 J =240 g.°C×c
c = 48 J/240 g.°C
c =0.2 J/g.°C
Answer:
Geologists commonly consider faults to be active if there has been movement observed or evidence of seismic activity during the last 10,000 years. Active faulting is considered to be a geologic hazard - one related to earthquakes as a cause.
Explanation:
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