Calculate the amount of heat (in J) required to heat 2.00 grams of ice from - 12.0 °C to 88.0 °C.

The heat of fusion of water is 6.02 kJ/mol, the specific heat capacity of ice is 2.09 J/g ▪°C and the specific heat capacity of water is 4.184 J/g ▪°C.

Question

Dayanara Torres

Physics

13 February, 06:04

Calculate the amount of heat (in J) required to heat 2.00 grams of ice from - 12.0 °C to 88.0 °C.

The heat of fusion of water is 6.02 kJ/mol, the specific heat capacity of ice is 2.09 J/g ▪°C and the specific heat capacity of water is 4.184 J/g ▪°C.

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Answers (1)

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Killian Waller · Answer 1

This heating process will be divided into three parts:

1) Heating of ice from - 12 °C to 0 °C

2) Melting of ice into water

3) Heating of water from 0 °C to 88 °C

Total heat = H (ice) + H (melting) + H (water)

Total heat = mCp (ice) ΔT (ice) + nl (fusion) + mCp (water) ΔT (water)

= 2 x 2.09 x (0 + 12) + 2/18 x 6020 + 2 x 4.184 x (88 - 0)

= 1455.4 Joules

Calculate the amount of heat (in J) required to heat 2.00 grams of ice from - 12.0 °C to 88.0 °C.The heat of fusion of water is 6.02 kJ/mol, the specific heat capacity of ice is 2.09 J/g ▪°C and the specific heat capacity of water is 4.184 J/g ▪°C.

Calculate the amount of heat (in J) required to heat 2.00 grams of ice from - 12.0 °C to 88.0 °C.

The heat of fusion of water is 6.02 kJ/mol, the specific heat capacity of ice is 2.09 J/g ▪°C and the specific heat capacity of water is 4.184 J/g ▪°C.