Answer:
Mass of substance m = 0.050Kg
Initial Temperature T₁ = 20°C
Final Temperature T₂ = 100°C
we have to calculate the heat absórbed by the lead .
Since spècific heat of lead is not gíven so we will take spècific heat of lead 128 J/kg°C
[tex] \diamond \bf \: Q\: = m \times c( \Delta \: t) [/tex]Put the given value
[tex]\sf \longrightarrow \: Q = 0.050 \times 128 \times (100 - 20) \\ \\ \\ \sf \longrightarrow \: Q = 0.050 \times 128 \times (80) \\ \\ \\ \sf \longrightarrow \: Q = 6.4 \times 80 \\ \\ \\ \sf \longrightarrow \: Q = 512 \: joules[/tex]
Therefore,
- Heat absórbed by the lead is 512 Joules.
