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.

Explanation:

Given: mass of lead = 4.64 kg

Convert kg into grams as follows.

$1 kg = 1000 g\\4.64 kg = 4.64 kg \times \frac{1000 g}{1 kg}\\= 4640 g$

$T_{1} = 150^{o}C$

$T_{2} = 219^{o}C$

The standard value of specific heat of lead is $0.128 J/g^{o}C$.

Formula used to calculate heat is as follows.

$q = m \times C \times \Delta T$

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.

$q = m \times C \times \Delta T\\= 4640 g \times 0.128 J/g^{o}C \times (219 - 150)^{o}C\\= 40980.48 J/g^{o}C$

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.