Specific latent heat of fusion lab

[Boogeyman]

Homework Statement

It's part of a lab designed to find the specific latent heat of fusion of ice using the method of mixtures. The results are in the 1st attachment..I want to find the specific latent heat of fusion of ice from those nos.

Homework Equations

heat lost by water = heat gained by ice + heat required to melt ice

The Attempt at a Solution

m_wc_w($$\theta_i$$) - ($$\theta_f$$ - m_ic_i($$\theta_i$$ - 0) + m_il_f

l_f = m_wc_w ($$\theta_i$$ - $$\theta_f$$) - m_ic_i($$\theta_i$$ - 0) / m_i

I don't know how to fix this cursed code...anyways this is the equation I come up with. My friend in his attempt, (2nd attachment), got something different. I also found if you used c_w instead of c_i at the end of the last eqn you got the precise value of the specific latent heat of fusion of ice.

 

[rl.bhat]

I also found if you used cw instead of ci at the end of the last eqn you got the precise value of the specific latent heat of fusion of ice.

If the ice is at zero degree celsius you can use cw in place of ci.

 

[Boogeyman]

If the ice is at zero degree celsius you can use cw in place of ci.

Ok could you explain why that is so, because as far as I know you would get different answers..btw when I worked it out yesterday I got 4.85x10⁵Jkg^-1

This is what, like more than 100 000 off the correct answer. Now I didn't use any insulation material during the experiment..can my answer be off by so much because of heat lost to surroundings?

 

[rl.bhat]

You are doing the experiment in a beaker. So you have to take into account the heat gained by the beaker also. The spacific heat of the glass is 840 J/kg .degree. Then the answer is still more. So you have made some error in the observations. Do it again.