specific heat

[intothevoidx]

If you want to use 3500 kilocalories of energy to heat up ice using body temperature how much water would you need? I got 22.5 liters, and I'm not sure if it's right or not. Your help is appreciated

 

 

thanks

[woelen]

It does not depend on body temperature, the only thing which counts is amount of energy.

 

You must be more specific with your question. What is the starting temperature of the ice, and to which temperature must the ice be heated?

 

What you need, is the specific heat of ice, the amount of energy, needed to melt a certain amount of ice to water of the same temperature, and the specific heat of water.

[intothevoidx]

The ice is starting at 0 degrees celsius and it is going to be heated to 37 degrees celsius. I was told to use the specific heat of water for ice.....unless that's wrong? 4.186......

[intothevoidx]

Can't I just use the equation for specific heat to find mass and convert it into liters?

[swansont]

You have to find the energy used to melt the ice before it warms up (latent heat) in addition to the energy used to warm it up.

[intothevoidx]

Well, then I'm out of luck, as this is my first year of chemistry, and frankly we're only two chapters in, unless someone on here would like to tutor me in their free time then I have no idea what to do lol.

[swansont]

It's just a second term in the equation. Here's some reading:

 

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase.html#c1

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/phase2.html#c1

[woelen]

Two numbers are needed to solve the equation. First, you need the amount of energy, needed for melting 1 kg of ice into 1 kg of water, both at 0 degrees. Next, you need the amount of energy, needed for heating 1 kilogram of water with 1 degree.

 

Amount of energy, needed for melting ice to water: 334 kJ/kg

Amount of energy, needed for heating liquid water with 1 degree: 4.186 kJ/kg/C

 

Now, suppose you have N kg of ice.

You want to heat N kilograms of ice to N kilograms of water of 37 degrees. Total amount of energy needed is

 

N*334 + N*4.186*37 kJ = N*(334 + 4.186*37) = 488.9*N

 

You have 3500 kJ of energy, so 488.9*N = 3500. So, N = 7.16 kg.

 

With 3500 kJ of energy, you can melt and heat 7.16 kg of ice, with an initial temperature of 0 degrees to a temperature of 37 degrees, provided that there are no losses.