Question

When 400 Joule of heat is given to 100 g sample of a metal, its temperature increased by 20°C. If specific heat of metal is n × 50 J kg1°C1 then find the value of n ?​

Answer 1

Answer:

Text Solution. Solution : Given, Heat ΔQ = 400 J, Mass m = 0.1kg. Temperature ΔT = 20∘C ∴ Specific heat, c=1 m.

Answer 2

Answer:

THE ANSWER IS 4.

Explanation:

N*50=400*100/20*10

N=200/50

N=4

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