College

A bag filled with lead shots is dropped from a height of [tex]h = 38.2 \, \text{m}[/tex]. The total mass of the bag is [tex]m = 745.0 \, \text{g}[/tex]. What is the increase in the temperature of the lead shots after the bag hits the ground? The specific heat of lead is [tex]c = 130 \, \text{J/kgK}[/tex].

Answer :

Answer:

The increase of temperature is 2.88 Kelvin

Explanation:

Step 1: Data given

Height = 38.2 meter

Total mass of the bag with lead m = 745.0 grams

Specific heat of lead c= 130J/kg* K

Step 2: Calculate change in temperature

Q = heat transfer

Q = m*c*ΔT

with m= mass

with c= specific heat capacity

with ΔT = change in temperature

Q = Potential energy

Potential energy = m*g*h

with m= mass

with g= acceleration due to gravity (9.81 m/s²)

with h= height


m*g*h = m*c*ΔT

ΔT = (m*g*h)/m*c

ΔT = g*h/c

ΔT = 9.81*38.2 /130

ΔT = 2.88 °C

The increase of temperature is 2.88 Kelvin

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