The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat

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The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat

er, and steam are 2.09 j/g−k, 4.18 j/g−k, and 1.84 j/g−k, respectively. for h2o, δ hfus = 6.01kj/mol, and δhvap = 40.67 kj/mol. the enthalpy change for converting 1.00 of ice at -50.0 to water at 60.0 is ________ . the specific heats of ice, water, and steam are 2.09 , 4.18 , and 1.84 , respectively. for , = 6.01, and = 40.67 . 12.28 6401 12.41 8.64 6.37

Chemistry

1 answer:

eduard [944] · Answer 1 · 2026-03-01T11:33:16+03:00

To begin with, we need to determine:

1. The heat required to raise the temperature of ice from -50 to 0 °C:

using the formula q:

q1 = m*C*ΔT

Where m is the mass of the ice, calculated as mol * molar mass

= 1 mol * 18 g/mol

= 18 g

and C is the specific heat of ice = 2.09 J/g-K

and ΔT is the temperature change = 0 - (-50) = 50°C

Thus, by substituting the values:

∴q1 = 18 g * 2.09 J/g-K * 50°C

= 1881 J = 1.881 KJ

2. The heat needed to melt this quantity of ice is:

q2 = n*ΔHfus

Here, n refers to the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

By substituting:

q2 = 1 mol * 6.01 KJ/mol

= 6.01 KJ

3. The heat necessary to raise the temperature of the water from 0°C to 60 °C is:

q3 = m*C*ΔT

Here, m is the mass of the water = 18 g

C represents the specific heat capacity of water = 4.18 J/g-K

ΔT signifies the temperature change for the water = 60°C - 0°C = 60°C

After substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

= 4514 J = 4.514 KJ

∴The overall change in enthalpy = q1 + q2 + q3

= 1.881 KJ + 6.01 KJ + 4.514 KJ

= 12.405 KJ