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Q: How much heat required for chainge temp of 28lbs of ice from -38 deg f to 32 deg f?

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Heat required for this transition is given as the the sum of three heatsheat required for heating the ice from -5 degree Celsius +latent heat(conversion of ice at zero degree to water at zero degrees)+heat required to heat the water from 0 to 5 degree CelsiusHeating of ice=m x s x delta T,where m is the mass ,s is the specific heat of ice=200x0.5x5=500calmelting of ice=mxlatent heat=200x80=16,000calHeating of water=m x s x delta T,where m is the mass ,s is the specific heat of water =200x1x5=1000calTotal heat required=500+16,000+1000=17,500 cal

question makes no sense.....

Assuming standard atmospheric pressure, 2260 kilojoules.

That completely depends on the temperature of the water when the gas flame is first ignited, which you've neglected to mention.

You mean how much heat energy will be lost/transferred as you are losing Joules here. All in steam, so a simple q problem and no change of state. 2.67 kg = 2670 grams q = (2670 grams steam)(2.0 J/gC)(105 C - 282 C) = - 9.45 X 105 Joules ----------------------------------- This much heat energy must be lost to lower the temperature of the steam.

Related questions

28lbs

28lbs, about as much as my 10 month old son.

The measurement of how much heat energy is required for a substance to melt is called the heat of fusion. It is the amount of energy required to change a substance from a solid to a liquid at its melting point.

How much heat (in calories) is required to heat a 43 g sample of aluminum from 72 F to 145F

The specific heat capacity of aluminum is 0.897 J/g°C. The heat required can be calculated using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values, you can calculate how much heat is required.

No, the amount of heat required to boil 1kg of water is much higher than the amount of heat required to melt 1kg of ice. Boiling water requires additional heat to overcome the latent heat of vaporization, while melting ice only requires heat to overcome the latent heat of fusion.

lf = 3.35 x 105 J kg-1 This much amount of heat required to convert 1 kg of ice to liquid Mani.Ra

The heat required to convert ice at 0°C to water at 0°C is known as the latent heat of fusion. For water, this value is 334 J/g. Therefore, to convert 0.3 g of ice to water at the same temperature, the heat required is 0.3 g * 334 J/g = 100.2 Joules.

The values are different for each type of plastic.

No heat (energy) is required to freeze water (from liquid to solid). Freezing RELEASES energy (heat), as it is an exothermic event. If you want to know how much energy is release, you need to know the heat of fusion for water, and then multiply that by the mass of water being frozen.

To calculate the heat energy required, you can use the formula: Q = mcΔT, where Q is the heat energy, m is the mass of the copper (0.365 kg), c is the specific heat capacity of copper (0.0920 J/g°C), and ΔT is the change in temperature (60.0°C - 23.0°C). First, convert the mass to grams and then plug the values into the formula to find the heat energy required.

The specific heat capacity of water is 4186 J/kg*C. To calculate the heat required, use the formula: heat = mass * specific heat capacity * change in temperature. Plugging in the values, the heat required to raise the temperature of 0.25 kg of water by 10 degrees Celsius is approximately 1046.5 Joules.