Heat Transfer Question

hey i calculate it to be 627,000 joules because

it takes 1 calorie to heat 1 gram by 1 degree and theres 1500 grams so thats going to take 1500 calories to heat the whole 1.5kg's by 1 degree so if you multiply the 1500 calories by 100 because you want to reach 100 celsius that gives you an answer of 150,000 and thats how many calories it will take and there are 4.18 joules in 1 calorie so multiple 150,000 by 4.18 which gives you 627,000 joules

WRONG!

You left out the latent heat at both ends!

The latent heat of fusion required to convert 1.5 kg from ice at 0°C to water at 0°C is 501 kJ (501,000 Joules).

The heat required to raise the temperature of 1.5 kg of liquid water from 0°C to 100°C is 627 kJ (The only part you got right)

The latent heat of vaporization required to convert 1.5 kg of water at 100°C to steam at 100°C is 3390 kJ.

Summing the total amount of heat energy required yields the correct answer of:

4518 kJ (4,518,000 Joules)

For some insight:

http://en.wikipedia.org/wiki/Latent_heat

Hi, Your task is to heat 1.5kg of ice from 0ºC to steam 100ºC. Since the product in question is water and it will exhibit all the three states of matter during the thermodynamic process, you need to supply heat for: i) Fusion (latent Heat of Fusion at 0ºC). ii). Boiling from 0-100ºC and iii). Vapourization (Latent Heat of Vapourization at 100ºC).

Mathematically, the equation can be thus summarized:

Q = mþ + mc_p(100-0) + mß OR m{þ + c_p(100-0) + ß}

where m = 1.5kg, the mass of water

þ = 336,000 J/kg, specific latent heat of fusion of ice

ß = 2,290,000 J/kg, specific latent heat of steam

c_p= Specific Heat Capacity of H₂O ( this quantity is temperature dependent in the following order....c_p = A + BT +CT² + DT³ + ET⁴ + ......but can be assumed as constant for our case. Therefore c_p = 4200 J/kg. the heat required to do this job is therefore 1.5{336,000 + 4200(100-0) + 2,290,000) Joules, thet is 4.57 MJ (Megajoules)

You are correct that the specific heat of water varies with temperature. In our range (0-100°C) due to the unique properties of water, it first declines then rises again as shown:

(I should have added "per °C" to my graph title and y-axis label. Also the first data point at 0°C is 4217 J/kg but did not copy over)

An average of these values yields 419 kJ/kg, falling midway between the value I used and the one you used.

The values you used for the latent heats are higher than the ones I used and I do not know where you obtained them but they are higher than the values from any sources I have checked.

For latent heat of fusion (melting) the number centers at the 334 kJ/kg I used, whereas you are using 336.

For the latent heat of vaporization (evaporation) I used 2260 kJ/kg, but upon rechecking other sources I did find 2270, but nothing higher, whereas your value is 2290.

Recalculating, using the highest values I found, yields a total heat of 4534.5 kJ which rounded to the significant digits = 4530 kJ (4.53 MJ) for the 1.5 kg of ice converted to steam. That is less than 0.4% higher than my first answer and about 0.8% lower than yours. While we were within about 1% of each other, I am curious as to where you got your latent heat values since they differ significantly from anything I have come across. Having encountered similar small, yet significant discrepancies in a variety of values obtained from different sources many times in my career, I am not prepared to say you are wrong, but I feel compelled to question their source.

The main issue was addressed by us both and that is simply the latent heat in the transition form ice to water, and water to steam.

Regards,

Greg

http://www.simetric.co.uk/si_steam.htm

You guys! stop doing this kid's homework for him/her.

Shame on you that posted answers.

Good job to those that promoted his/her research by pointing to steam tables, equations and constants.

By the way OP, knowing your question is more important than finding the answer. (why? because if you don't ask the right question, you're probably not going to get the right answer.) This is not a Heat Transfer question, this is Thermodynamics.

The answer to your question lies in the equations of thermodynamics... not heat transfer. Refer to your thermo books on this one.

Correct! Thermodynamics is not concerned with the rate at which processes take place, whereas heat transfer is mainly about the rate.