Answer:
The right option is (d).148 cal/g
Step-by-step explanation:
Specific Latent Heat Of Vaporization: This is the quantity of heat required to change unit mass of a substance from either liquid to vapor, or solid to vapor without a change in temperature.
Heat gained by dry ice = Heat lost by the water
Q₁ = Q₂................. equation 1
Where Q₁ = heat gained by the dry ice, Q₂ = heat lost by the water.
Q₁ = lm₁........................ equation 2
Q₂ = cm₂ΔT................... equation 3
Substituting the equation 1 into Equation 2,
lm₁ = cm₂ΔT ................... equation 4
Where l = heat of vaporization of the dry ice, m₁ = mass of the dry ice, c = specific heat capacity of water, m₂ = mass of water, ΔT = change in temperature = T₁ - T₂
Making l the subject of formula in equation 4
l = cm₂ΔT/m₁...................... equation 5
Where c = 1.00 cal/g.°C, m₂ = 500g, ΔT = T₁ - T₂ = 66-29 = 37°C, m₁ = 125 g.
Substituting these values into equation 5
∴ l = (1 × 500 × 37)/125
l = 148 cal/g
The right option is (d).148 cal/g