Which has more heat 1g of ice or 1g of water?
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Which has more heat 1g of ice or 1g of water?
1 g of water at 0°C has more heat than 1 g of ice at 0°C. This is because ice at 0°C absorbs 360 J of heat energy to convert into water at 0°C.
How many calories of heat will be required to convert 1 gram of ice at zero degree Celsius into steam at 100 degree Celsius?
Total heat required to convert 1 g of ice at 0°C into steam at 100°C is 716 cal.
Which requires more heat 1g of ice?
1 gram of ice at 0°C requires additional 80 calories of heat to get converted into water at 0°C. Then, heat is provided to raise the temperature to 10°C. Therefore, ice requires more heat than water and the additional heat is known as ‘Latent heat of fusion of ice’.
Is the amount of energy that it takes to raise the temperature of 1 g of a substance by 1 Kelvin?
The specific heat is the amount of heat required to raise the temperature of 1 g of substance by one degree Celsius or one Kelvin.
How much energy does it take to convert ice to water?
Converting 100. g of ice at 0.00 °C to water vapour at 100.00 °C requires 301 kJ of energy. There are three heats to consider: q1 = heat required to melt the ice to water at 0.00 °C. q2 = heat required to warm the water from 0.00 °C to 100.00 °C.
How many joules of heat does ice have?
45958 joules. c is the specific heat capacity of the substance, which is 4.186 J g ∘C for water But then, we also need to change the ice to water first, and so we can use the latent heat formula, which states that,
How do you calculate the total energy required to heat ice?
The total energy required is the sum of the energy to heat the -10 °C ice to 0 °C ice, melting the 0 °C ice into 0 °C water, heating the water to 100 °C, converting 100 °C water to 100 °C steam and heating the steam to 150 °C. To get the final value, first calculate the individual energy values and then add them up.
What is the specific heat of steam and ice?
specific heat of ice = 2.09 J/g·°C. specific heat of water = 4.18 J/g·°C. specific heat of steam = 2.09 J/g·°C. Solution: The total energy required is the sum of the energy to heat the -10 °C ice to 0 °C ice, melting the 0 °C ice into 0 °C water, heating the water to 100 °C, converting 100 °C water to 100 °C steam and heating the steam to 150 °C.