A bag filled with lead shots is dropped from a height of h = 38.2 m. The total mass of the bag is m = 745.0 g. What is the increase in the temperature of the lead shots, after the bag hits the ground? The specific heat of lead is c = 130 J/kgK. Tries 0/99

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