Respuesta :
Answer : The number of ice cubes melt must be, 13
Explanation :
First we have to calculate the mass of water.
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}[/tex]
Density of water = 1.00 g/mL
Volume of water = 650 mL
[tex]\text{Mass of water}=1.00g/mL\times 650mL=650g[/tex]
Now we have to calculate the heat released on cooling.
Heat released on cooling = [tex]m\times c\times (T_2-T_1)[/tex]
where,
m = mass of water = 650 g
c = specific heat capacity of water = [tex]4.18J/g^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]29^oC[/tex]
[tex]T_2[/tex] = initial temperature = [tex]0^oC[/tex]
Now put all the given values in the above expression, we get:
Heat released on cooling = [tex]650g\times 4.18J/g^oC\times (29-0)^oC[/tex]
Heat released on cooling = 78793 J = 78.793 kJ (1 J = 0.001 kJ)
As, 1 ice cube contains 1 mole of water.
The heat required for 1 ice cube to melt = 6.02 kJ
Now we have to calculate the number of ice cubes melted.
Number of ice cubes melted = [tex]\frac{\text{Total heat}}{\text{Heat for 1 ice cube}}[/tex]
Number of ice cubes melted = [tex]\frac{78.793kJ}{6.02kJ}[/tex]
Number of ice cubes melted = 13.1 ≈ 13
Therefore, the number of ice cubes melt must be, 13
We have that for the Question "Approximately how many ice cubes must melt to cool 650 milliliters of water from 29°C to 0°C?"
- [tex]14 icecubes[/tex]
From the question we are told
Assume that each ice cube contains 1 mole of H2O and is initially at 0°C. ∆H(fusion) = 6.02 kJ/mol; ∆H(vaporization) = 40.7 kJ/mol c(solid) = 2.09 J/g°C; c(liquid) = 4.18 J/g°C; c(gas) = 1.97 J/g°C.
Generally the equation for mass is given as
[tex]Mass = D*V\\\\=1g/ml * 650\\\\ = 650g[/tex]
where:
[tex]\{dH} = m C_liq \{dT}[/tex]
[tex]= 650*4.18*(29-0)\\\\= 78793J or 78.79KJ[/tex]
[tex]\{dH_ice = 6.02 kJ/mol\\\\Moles of ice = \frac{78.79}{6.02}[/tex]
[tex]=13.08 approximately 14 ice cubes[/tex]
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