Question

1. How much heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C?

Answer 1

Answer: An amount of
`40980.48 J/g^(o)C` heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.

Step-by-step explanation:

Given: mass of lead = 4.64 kg

Convert kg into grams as follows.

`1 kg = 1000 g\\4.64 kg = 4.64 kg * (1000 g)/(1 kg)\\= 4640 g`

`T_(1) = 150^(o)C`

`T_(2) = 219^(o)C`

The standard value of specific heat of lead is
`0.128 J/g^(o)C`.

Formula used to calculate heat is as follows.

`q = m * C * \Delta T`

where,

q = heat energy

m = mass of substance

C = specific heat of substance

`\Delta T` = change in temperature

Substitute the value into above formula as follows.

`q = m * C * \Delta T\\= 4640 g * 0.128 J/g^(o)C * (219 - 150)^(o)C\\= 40980.48 J/g^(o)C`

Thus, we can conclude that
`40980.48 J/g^(o)C` heat is required to raise the temperature of 4.64kg of lead from 150°C to 219°C.