Question

360J of heat is supplied to rise the temperature of a a certain mass of iron from 20°C To 25 °C‐1. if the specific heat of iron is 470Jkg-1°C-1, then the mass of iron is​

Answer 1

Explanation:

Q= mcΔΦ

where Φ is temperature

360= m × 470 × (25-20)

360= m× 470×5

360= m × 2350

m=360/2350

m = 0.153kg

Answer 2

Answer:

360J=m×470(5)

360j=m×2350

m=360/2350

m=0.153kg