Question

How much energy is required to raise the temperature of a 300.0

gram block of lead from 22.3°C to 59.9°C? The specific heat of lead is

0.129 J/gºC.

Answer 1

Q = 1455.12 Joules.

Step-by-step explanation:

Given the following data;

Mass = 300 grams

Initial temperature = 22.3

Final temperature = 59.9°C

Specific heat capacity = 0.129 J/gºC.

To find the quantity of energy;

`Q = mcdt`

Where,

Q represents the heat capacity.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt represents the change in temperature.

dt = T2 - T1

dt = 59.9 - 22.3

dt = 37.6°C

Substituting the values into the equation, we have;

`Q = 300*0.129*37.6`

Q = 1455.12 Joules.