Question

A piece of lead falls from a height of 100 m on a fixed non-conducting slab which brings it to rest. If the specific heat of lead is 30.6 cal/g the increase in temperature of the slab immadiatedly after collision

Answer 1

The amount of heat evolved during the reaction or process is equal to the product of specific heat, mass of mixture and temperature change during reaction. So, it can be written as follows:

Heat evolved (q) = specific heat (C) x Mass of mixture (m) x Temperature change (ΔT)

Mass of lead is 75.0 g.

The temperature change is 10° C

Heat evolved in the process is 96 J

The formula q=C x m x ΔT also is written as follows:

C= q/(m x ΔT)

Substitute all required values in the equation:

C= 96 J/(75 g x 10° C)

= 0.128 J/g* °C

Therefore, the specific heat of lead is 0.128 J/g* °C

Answer 2

Answer:

7.62 c

Explanation:

gravity = 9.8

height = 100m

joule = 4.8

specific heat = 30.6 cal/kg

$t = \frac{gh}{js}$

where as t = increase in temperature

$t = \frac{9.8 \times 100}{4.8 \times 30.6}$

$t = 7.62 \: c$

I hope the answer will help you

thank you