Question

A 500 gm cube of lead is heated from 25°C to 75°C. How much energy was required to heat the lead? The specific heat of lead is 0.129 J/9°C.

Give proper steps while solving this please​

Answer 1

Given:

• Mass of lead cube is 500g
• $\bold{T_{1}}$ = 25°
• $\bold{T_{2}}$ = 75°
• Specific heat of land is 0.129 J/g°C

Formula used:

$\pink{\fbox{Q = (mass) (∆T) (Cp)}}$

Answer:

Here, ∆T = $\bold{T_{2}}$ - $\bold{T_{1}}$

= 75 - 50

= 50

On substituting the values,

Q = (500) (50) (0.129)

Q = 3225 joules

Hence, the required energy is 3225 joules.

_____________________________