(1 ~kg ) of steam at (150^ C ) is passed from a steam chamber is to a copper coil immersed in (20 ~L ) of water. The steam condenses in the coil and is returned to the steam chamber as water at (90^ C ). Latent heat of steam is (540 cal g ⁻¹ ), specific heat of the steam is (1 cal ) and ( g ^ -1 ^ C ⁻¹ ). Then, the rise in temperature of water is

Step-by-step solution

Mass of steam, (m_s=1 ~kg =1000 ~g =10^3 ) Temperature of steam, (T₁=150^ C ) Latent heat of steam, ( aligned L_s & =540 cal g ⁻¹ c & =1 cal g ^ -1 ^ C ⁻¹ aligned ) Heat lost by steam, ( aligned Q^ & =m L_f+m_s c t & =10^3 540+10^3 1 (150-90) & =54 10^4+6 10^4 & =10^4(54+6)=60 10^4 cal aligned ) Heat gained by (20 ~L ) water, (Q^ =m_w c T ) Mass of (20 ~L ) water, (m_w=20 ~kg =20 10^3 ~g ) ( Q^ =20 10^3 1 T ) By the principle calorimetry, Heat lost (= ) Heat gained (60 10^4=20 10^3 T ) ( T= 60 10^4 20 10^3 =30^ C )