How much heat is required to raise 36 g ice at – 10.0oC to steam at 110oC? (get your answers from question #1) a) Amount of heat to go from –10.0oC to 0.00oC (solid, no phase change)

The total heat required to raise ice at -10°C to steam at 110°C = 91606.8 J

Explanation:

The heat involved in this process involves the following:

1. Heat to change ice at -10°C to ice at 0°C;

2. Heat to change ice at 0°C to water at 0°C

3. Heat to change water at 0°C to water at 100°C

4. Heat to change water at 100°C to steam at 100°C

5. Heat to change steam at 100°C to steam at 110°C

Specific heat capacity of ice, c = 2040 J/K/kg, Latent heat of fusion of ice, L = 3.35 × 10⁵ J/kg, specific heat capacity of water, c =  4182 J/K/kg, latent heat of vaporization of water, l = 2.26 × 10⁶ J/kg, specific heat capacity of steam, c = 1996 J/K/kg

Step 1: H = mcθ; where m = 30.0 g = 0.03 g, c = 2040 J/K/kg, θ = (0 -  -10) = 10 K

H = 0.03 * 2040 * 10 = 612 J

Step 2: H = mL, where  L = 3.35 × 10⁵ J/kg

H = 0.03 * 3.35 × 10⁵ = 10050 J

Step 3: H = mcθ, where c =  4182 J/K/kg, θ = (100 - 0) = 100 K

H = 0.03 * 4182 * 100 = 12546 J

Step 4: H = ml, where l = 2.26 × 10⁶

H = 0.03 * 2.26 × 10⁶ = 67800 J

Step 5: H = mcθ, where c = 1996 J/K/kg, θ = (110 - 100) = 10 K

H = 0.03 * 1996 * 10 = 598.8 J

Total heat required to raise ice at -10°C to steam at 110°C = (612 + 10050 + 12546 + 67800 + 598.8) J = 91606.8 J

Therefore, the total heat required to raise ice at -10°C to steam at 110°C = 91606.8 J

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