Question

A 10.0kg – iron ball is dropped onto the pavement from a height of 500.0m. If half the heat generated goes into warming the ball, find the temperature increase of the ball. (The specific capacity for iron is 450J/kg·C & g=10.0m/s2). It is assumed your answer will be in degrees centigrade so just enter a number.(Example: 15.2C=15.2)

Answer 1

The temperature increase of the 10.0kg iron ball, after being dropped from a height of 500.0m and absorbing half of the heat generated, is approximately 5.56°C.

Step-by-step explanation:

To find the temperature increase of the 10.0kg iron ball after being dropped from a height of 500.0m, we must first determine the amount of potential energy converted to heat as the ball hits the pavement. The potential energy (PE) of an object is given by PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. Using m = 10.0 kg, g = 10.0 m/s2, and h = 500.0 m, we get:

PE = (10.0 kg) × (10.0 m/s2) × (500.0 m) = 50,000 joules

Since half the heat generated goes into warming the ball, the heat (Q) that goes into the ball is:

Q = 0.5 × 50,000 J = 25,000 J

Using the formula for heat transfer Q = mcΔT, where m is mass, c is the specific heat capacity, and ΔT is the change in temperature, and rearranging for ΔT, we get:

ΔT = Q / (mc)

Substituting in the values, ΔT = 25,000 J / (10.0 kg × 450 J/kg·°C) = 25,000 J / 4,500 J/°C = 5.56°C