The current in the heater is 3.8A and the potential difference (p.d.) across it is 12V. The iron block has a mass of 2.0 kg. When the heater is switched on for 10 minutes, the temperature of the block rises from 25*C to 55*C.
Calculate the specific heat capacity of iron.
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Answered by
4
Answer:
Explanation: I=3.8A, V=12, power,P=IV=w/t=45.6W, t=10min=600s
work,Q=pt
=45.6W×600s
=45.6J/s × 600s
=27360J
Δθ=change in temperatures=55°-25°=30°c=303k
Q=mcΔθ where c= specific heat capacity= S.H.S , m=mass
⇒c=Q/(mΔθ)
=27360/(2×303)
=27360/606
c =45.1Jkg⁻¹k⁻₁
Answered by
1
Answer:
» c=Q/(mA0)
=27360/(2*303)
=27360/606
c =45.1Jkg-k;
Explanation:
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