Question

A 960.0 g iron meteor impacts the earth at a speed of 1,268.0 m/s. If its kinetic energy is entirely converted to heat of the meteorite, what will the resultant temperature rise be? (The specific heat for iron is 0.113 cal/g•K.)

Answer 1

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)

Answer 2

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 $= \frac{184.454.Cal}{108.48\frac{Cal}{K} } = \frac{184.454.K}{108.48} = 1.7 K$

ΔT = 1.7 K = (237 + 1.7)⁰C = 274.7 ⁰C

Therefore, the resultant temperature rise is 274.7 ⁰C