Question

How much heat energy is required to raise the temperature of a 500g cube of lead from 25°C to 75°C ? (Specific heat capacity of lead is 0.13J/g.C)​

Answer 1

Answer:

• The Heat energy (Q) required is 3.250 KJ

Given:

1. Mass of cube = 500 grams.
2. Initial temperature (T₀) = 25°C
3. Final Temperature (T) = 75°C
4. Specific heat capacity = 0.13 J/g C

Explanation:

$\rule{300}{1.5}$

From the formula we know that,

$\bigstar\;\underline{\boxed{\sf Q=MC\;\Delta T}}$

Here,

• Q Denotes Heat energy.
• M Denotes Mass.
• C Denotes specific heat capacity.
• ΔT Denotes Temperature difference.

Solving and substituting the values,

$\longrightarrow\sf Q=M\;C\;\bigg(T-T_{o}\bigg)\\\\\\\\\longrightarrow\sf Q=500\times 0.13\times \bigg(75-25\bigg)\\\\\\\\\longrightarrow\sf Q=500\times 0.13 \times 50\\\\\\\\\longrightarrow\sf Q=25000\times 0.13\\\\\\\\\longrightarrow\sf Q=3250\\\\\\\\\longrightarrow\sf Q=3.250\times 10^{3}\\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf Q=3.250\;KJ}}}}$

∴ The Heat energy (Q) required is 3.250 KJ.

$\rule{300}{1.5}$

Answer 2

Answer:

The Heat energy (Q) required is 3.250 KJ

Given:

Mass of cube = 500 grams.

Initial temperature (T₀) = 25°C

Final Temperature (T) = 75°C

Specific heat capacity = 0.13 J/g C