Question

1. For each of your three trials state the following:
• heat needed to melt the ice (q)
• enthalpy of fusion
• percent error from the accepted enthalpy of fusion of water of 334 J/g

Answer 1

79.72 cal/gm is the Heat of fusion

Answer 2

) To heat 1 kg (about 1 liter) of water from 283.15 K to 303.15 K (10 °C to 30 °C) requires 83.6 kJ. However, to melt ice also requires energy. To heat ice from 273.15 K to water at 293.15 K (0 °C to 20 °C) requires:

(1) 333.55 J/g (heat of fusion of ice) = 333.55 kJ/kg = 333.55 kJ for 1 kg of ice to melt
PLUS
(2) 4.18 J/(g·K) × 20K = 4.18 kJ/(kg·K) × 20K = 83.6 kJ for 1 kg of water to increase in temperature by 20 K
= 417.15 kJ

Answer 3

@magicalsd

Answer 4

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