How much heat (in kJ) is required to warm 10.0 g of ice, initially at - 10.0 °C, to steam at 110.0 °C? The heat capacity of ice is 2.09 J/g · °C, and that of steam is 2.01 J/g °C.

30.5 kJ.

Explanation:

• how much heat is required to convert that sample of ice from solid at - 10.0°C to solid at 0°C.

• determine how much heat is required to convert that sample of ice from solid at 0°C to liquid at 0°C.

• how much heat is required to heat that sample of now liquid water from 0°C to 100.0°C.

• how much heat is required to heat that sample of now liquid water from 100°C to 110.0°C

q = m * Cp * Delta T;

Where,

q = Heat

m = mass

Cp = specific heat capacity

Delta T = change in temperature

1. Heat absorbed by the ice = 10 * 2.09 * (0 - (-10))

= 209 J

2. q = n * Heat of fusion

Where,

n = moles of substance

Number of moles of ice = mass/molar mass of water

Molar mass of water = (2*1) + 16

= 18 g/mol

Number of moles = 10/18

= 0.56 mol.

Heat of fusion = 6.02 kJ/mol

= 6.02 * 0.56

= 3.344 kJ

3. q = m * Cp * Delta T;

Cp of water = 4.18 J/g. K

= 10 * 4.18 * (100 - 0)

= 4180 J

4. q = n * Heat of vapourisation

Heat of vapourisation = 40.7 kJ/mol

= 0.56 * 40.7

= 22.6 kJ

5. q = m * Cp * Delta T;

Cp = 2.01 J/g. K

= 10 * 2.01 * (110 - 100)

= 201 J

= 209 J + 3344 J + 4180 J + 22600 J + 201

= 30.534 kJ

30.534 KJ to warm 10 grams of ice initially at - 10.0 degrees celsius, to steam at 110 degrees celsius.

4.719 kJ

Explanation:

Q = m (L1∆T + L2T2)

m = 10 g

L1 = 2.09 J/g.°C

L2 = 2.01 J/g.°C

T2 = 110°C

T1 = - 10°C

∆T = T2 - T1 = 110 - (-10) = 120°C

Q = 10 (2.09*120 + 2.01*110) = 10 (250.8 + 221.1) = 10 (471.9) = 4719 J = 4719/1000 = 4.719 kJ