Question 3(i)10125 J of heat energy boils off 4.5 g of water at 100°C to steam at 100°C. Find the specific latent heat of steam.

Question

Accepted Answer

Given,Mass (m) = 4.5 g = 4.51000\dfrac{4.5}{1000}10004.5​ = 4.5 x 10-3 kgHeat energy given out (Q) = 10,125 JSpecific latent heat of steam = ?From relation Q = m x LSubstituting the values in the formula we get,10125=4.5×10−3×L⇒L=101254.5×10−3⇒L=2.25×106 J kg−110125 = 4.5 \times 10^{-3} \times L \[0.5em] \Rightarrow L = \dfrac{10125}{4.5 \times 10^{-3}} \[0.5em] \Rightarrow L = 2.25 \times 10^{6} \text{ J kg}^{-1} \[0.5em]10125=4.5×10−3×L⇒L=4.5×10−310125​⇒L=2.25×106 J kg−1Hence, the specific latent heat of steam = 2.25 x 106 J kg-1

Solved Sample Paper 1 — Question 23

Question 3(i)