Question

The steel ball with a volume of 3.6 cm³ and a mass of 28 g is heated from 20 °C to 100 °C. Calculate the thermal energy gained by the ball. (Specific heat capacity of steel = 510 J/kg °C)

a) 48.24 J
b) 56.7 J
c) 63.36 J
d) 72.9 J

Answer 1

The thermal energy gained by the steel ball can be calculated using the formula Q = m * c * ΔT, where Q is the thermal energy gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The thermal energy gained by the ball is 112,560 J (option d).

Step-by-step explanation:

The thermal energy gained by the steel ball can be calculated using the formula:

Q = m * c * ΔT

where Q is the thermal energy gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

Substituting the given values into the formula:

Q = 28g * (510 J/kg °C) * (100 °C - 20 °C)

Q = 28g * 510 J/kg °C * 80 °C

Q = 112,560 J

Therefore, the thermal energy gained by the ball is 112,560 J (option d).