- #1

fluidistic

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## Homework Statement

Ok, this is something related to a real life problem, more precisely a laboratory experiment.

Say that the temperature of the room is 25°C. I have a calorimeter. I chose to put 250 g of water at 27.5 °C inside it. But as it is not a perfect calorimeter, say I've calculated (by the least squares method and by doing a long experiment) its equivalent mass in water to be 50 g. In other words, the calorimeter behaves like absorbing the same heat as 50 g of water.

Now I want to put a 20 g ice cube in order to calculate the latent heat of water.

To find it, I think of using the formula [tex]Q_1+Q_2=0[/tex]. Or, in details : [tex](M+\pi)c_{\text{water}}(T_f-27.5)+mc_{\text{ice}}(0-T_i)+mc_{\text{water}}(T_f-0°C)+mL_{\text{ice}}=0[/tex] where [tex]T_i[/tex] is the initial temperature of the ice (which might be below 0°C), [tex]T_f[/tex] is the final temperature of the calorimeter and the melted ice cube, M is the initial mass of water, that is 250 g, [tex]\pi[/tex] is the equivalent mass in water of the calorimeter.

After measuring [tex]T_f[/tex], the only unknown in the equation would be [tex]L_{\text{ice}}[/tex].Ideally, what I would like is to have a [tex]\Delta T[/tex] such that the room temperature ([tex]T_r[/tex]) is such as [tex]T_r-\frac{\Delta T}{2}<T_r<T_r+\frac{\Delta T}{2}[/tex].

[tex]\Delta T[/tex] is the difference of temperature between the initial water+calorimeter temperature and the final temperature of the water+calorimeter+melted ice cube.

So ideally, in my example I would like to have [tex]T_f=[/tex]25°C-2.5°C=22.5°C, although I know it will likely not happen in my example.

I wrote all this to ask... am I making any error somewhere? I feel I'm right on everything, but I might miss something. Especially regarding the equation [tex](M+\pi)c_{\text{water}}(T_f-27.5)+mc_{\text{ice}}(0-T_i)+mc_{\text{water}}(T_f-0°C)+mL_{\text{ice}}=0[/tex].

Thank you infinitely.