Calculate the mass of steam at 100-C needed to just melt 320 g of ice at -10-C- The specific heat of ice -0-5 cal g-1-C-1- latent heat of vaporization of steam -540 cal g-1- specific heat of fusion of ice -80 cal g-1-

Calculate the amount of heat required to convert 1.00 kg of ice at -10°C at normal pressure to steam at 100°C. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹ , latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹ , specific heat capacity of water = 4200 J kg ⁻¹ K⁻¹ and latent heat of vaporization of water = 2.25 × 10⁶ J kg⁻¹

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Calculate the mass of steam at 100∘C needed to just melt 320 g of ice at −10∘C. The specific heat of ice =0.5calg−1∘C−1, latent heat of vaporization of steam =540calg−1, specific heat of fusion of ice =80calg−1.

Heat required =320×12×10+320×80=1600+25600=27200calLet the mas os steam required = mm×540=27200⇒m=50.370gm

Calculate the mass of steam at $100^{\circ} C$ needed to just melt 320 g of ice at $- 10^{\circ} C$ . The specific heat of ice $= 0.5 c a l g^{- 1}^{\circ} C^{- 1}$ , latent heat of vaporization of steam $= 540 c a l g^{- 1},$ specific heat of fusion of ice $= 80 c a l g^{- 1}$ .

Heat required $= 320 \times \frac{1}{2} \times 10 + 320 \times 80 = 1600 + 25600 = 27200 c a l$
Let the mas os steam required = $m$
$m \times 540 = 27200 \Rightarrow m = 50.370 g m$