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.

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

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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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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.