Answer:
The resultant temperature rise is 274.7 ⁰C
Explanation:
If the iron meteor kinetic energy is entirely converted to heat of the meteorite, then ΔKE = ΔThermal energy
ΔKE = ¹/₂ mv²
where;
m is mass in kg
v is velocity in m/s
ΔKE = ¹/₂ × 0.96 ×(1268)² = 771755.52 J = 184.454 cal
ΔThermal energy = mcΔT
where;
m is mass in g
c is the specific heat for iron = 0.113 cal/g•K
ΔT change in temperature or resultant temperature rise
Since, ΔKE = mcΔT
184.454 cal = 960 g × (0.113 cal/g•K) × ΔT
184.454 cal = (108.48 cal/K)ΔT
ΔT [tex]= \frac{184.454.Cal}{108.48\frac{Cal}{K} } = \frac{184.454.K}{108.48} = 1.7 K[/tex]
ΔT = 1.7 K = (237 + 1.7)⁰C = 274.7 ⁰C
Therefore, the resultant temperature rise is 274.7 ⁰C
Answer:
∆T = 1699.2 K = 1699.2°C
The resultant temperature rise is 1699.2 °C
Explanation:
The kinetic energy can be expressed as;
K.E = 0.5mv^2 ....1
Heat energy can be expressed as;
H.E = mc∆T .....2
Where;
m = mass of meteorite
v = speed of meteorite
c = specific heat capacity of iron
∆T = change in temperature.
If the kinetic energy is entirely converted to heat energy with no energy loss.
K.E = H.E
Substituting equation 1 and 2
0.5mv^2 = mc∆T
Making ∆T the subject of formula.
∆T = 0.5v^2/c .......3
Given;
v = 1268m/s
c = 0.113cal/gK × 4186.8J/kg per cal/g
c = 473.11J/Kg.K
Using equation 3, substituting the values
∆T = 0.5 × 1268^2 / 473.11
∆T = 1699.2 K = 1699.2°C
The resultant temperature rise is 1699.2 °C
Note: For temperature change, ∆T (Kelvin) = ∆T(°C)
