Lab 15:2 Heat of Fusion of Ice

[carmd]

Hi--

I just wanted to make sure that, in Table 15-2, p.275, G, where the heat of fusion of ice is calculated, I'm using the right equation. (I couldn't find it in the book or in the Answer Guide, but it's quite possible that I just missed it!)

The equation I'm going to use is
heat of fusion (in cal/g) = mass x c x delta t,

where
"mass" is the mass of the ice melted (calculated using F in the table),
"c" is the specific heat of water (1 cal/(g∙°C)) (not the speed of light )
and "delta t" is the temperature change (E in the table).

Thanks!

[pgarvan]

Assuming an ideal calorimeter for this experiment, the heat given up by the water must be exactly equal to the heat gained by the ice.

Now, the heat gained by the ice is used for two things: (i) melting the ice (i.e. fusion); and (ii) heating the melted ice to a higher temperature.

So, the math looks like this, assuming the specific heat of water is 1 cal/(g . degC)):

[Heat decrease in the water] = [Heat to melt the ice] + [Heat to warm the melted ice]

[m_w * c_w * dT_w] = [m_i * dH_f] + [m_i * c_w * dT_i]

where

m_w = mass of the water we started with in g
m_i = mass of melted ice in g
c_w = specific heat of water = 1 cal / g.degC
dT_w = decrease in the water temperature in degC
dT_i = increase in the ice temperature in degC = final temperature of the water
dH_f = heat of fusion of ice (what we're trying to find) in cal / g

This gives the following:

dH_f = c_w * (m_w * dT_w - m_i * dT_i) / m_i

Now, if the final temperature of the water is 0 degrees celsius, then dT_i = 0, and we get the following:

dH_f = c_w * (m_w * dT_w) / m_i