Question

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Question :

A metallic piece of 500g of lead is melted in a crucible and it is observed that a 40 watt electric heater will keep it in the liquid state at its melting point. If the heater is switched off, the temperature of the lead begins to fall after 5 min. calculate the specific latent heat of lead
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Answer 1

Heya!!

Here's your answer!!

Mass of the metal = 500g = 0.5 kg

Power of the electric heater = 40 watt = 40 J/s

Time for which current is passed = 5min = 5×60 = 300sec.

Specific heat capacity of metal = L

Heat energy supplied by heater = P×t =40×300

Heat energy radiated by metal = 0.5×L

0.5 L = 40×300

L = 40×300/0.5

= 24000 J kg-¹°C-¹.

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Answer 2

Answer:

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