Question

What is the amount of heat required to completely melt a 200.-gram sample of h2o(s) at stp?

Answer 1

The heat required to melt a 200-gram sample of ice at STP can be calculated using the heat of fusion, which is 79.9 cal/g for water. Multiplying this by 200 grams, the amount of heat required is 15,980 calories or 66,873.52 joules after conversion.

Step-by-step explanation:

Heat Required to Melt Ice at STP

To calculate the heat required to melt ice (solid H2O) at standard temperature and pressure (STP), we use the property known as the heat of fusion. The heat of fusion for water is the amount of energy needed to change 1 gram of ice at its melting point into 1 gram of liquid water without changing its temperature. According to data, this value is 79.9 cal/g for water. Therefore, to melt a 200-gram sample of ice, we would need:

Heat required = mass of ice × heat of fusion

= (200 g) × (79.9 cal/g)

= 15,980 calories

To convert calories to joules (since 1 calorie = 4.184 joules), the heat in joules is:

Heat required in joules = 15,980 cal × 4.184 J/cal

= 66,873.52 joules

Therefore, to completely melt a 200-gram sample of ice at STP, 66,873.52 joules of heat are required.

Answer 2

Answer: 66.7 kJ

Step-by-step explanation:


1) The amounf of heat to melt an amount of water (H₂O s) is equal to the amount of water times the melting latent heat.

2) The melting latent heat of water is found in tables or a textbook. It is 6.01 kJ / mol


3) Since, the melting latent heat of water found is per mol, we have to convert the givent mass of water into number of moles:

number of moles = mass in grams / molar mass = 200 g / 18.01 g/mol = 11.1 mol

4) Calculation of the heat required:

Heat = 11.1 moles × 6.01 kJ / mol = 66.7 kJ.