Question

A piece of iron block moves across a rough horizontal surface before coming to rest. The mass of the block is 3.3 kg, and its initial speed is 3.2 m/s. How much does the block's temperature increase, if it absorbs 74% of its initial kinetic energy as internal energy? The specific heat of iron is 452 J/(kg · °C).
°C

Answer 1

Answer:

Rise in temperature is given as

$\Delta T = 8.4 \times 10^{-3} ^0C$

Explanation:

Initial kinetic energy of the block is given as

$KE = \frac{1}{2}mv^2$

here we will have

m = 3.3 kg

v = 3.2 m/s

now we will have

$KE = \frac{1}{2}mv^2$

now we will have

$KE = \frac{1}{2}(3.3)(3.2)^2$

$KE = 17 J$

now we know that 74% of initial kinetic energy is absorbed as internal energy of the block

so the rise in temperature of the block is given as

$KE = ms\Delta T$

$0.74 \times 17 J = (3.3)(452)\Delta T$

$12.5 = 1491.6 \Delta T$

$\Delta T = 8.4 \times 10^{-3} ^0C$

Answer 2

Answer:

The temperature increases 0.0084ºC

Explanation:

Please look at the solution in the attached Word file.