A 25.0-g sample of ice at -6.5oC is removed from the freezer and allowed to warm until it melts. Given the data below, select all the options that correctly reflect the calculations needed to determine the total heat change for this process.

Melting point at 1 atm = 0,0°C; -2.09 J/g.°C; Cloud -4.21J/g °C, AH -6,02 kJ/mol
Check all that apply.

A. q for the temperature change from -6,5°C to 0.0°C is glven by 25.0 x 6.02 = 151 kJ
B. q for the phase change is given by 1.39 x 6.02 = 8.37 kJ
C. There are 3 separate heat change stages in this process.
D. The total heat change for the process is equal to +8.71 kJ
E. The total heat change for the process is equal to +350 kJ

B, D

Explanation:

The strategy here is to realize that the ice will be taken from -6.5 ºC to OºC where it will melt.

Lets call q₁ the heat required to bring the ice to 0ºc, q₂ the heat required to bring the phase change from solid to liquid.

q₁ is calculated from the expression

q₁ = s x m x ΔT where m is the mass, s the specific heat of ice ( 2.09 J/gºC ) and ΔT  the change in temperature.

q₂ the fusion enthalpy change   is calculated from the expression:

q₂ = C x ΔT

where C is the specific heat for the phase change , in this case named AH  given in kJ/mol.

We are given all the data needed to calculate q₁, q₂ and qtotal ( q₁ + q₂ )

q₁ = 25.0 g x ( 2.09 J/gºC) x ( 0 - ( -6.5 ºC ) )

q₁ = 339.6 J = 0.339 kJ

q₂ = (25 g/ 18 g/mol) x 6.02 kJ/mol = 1.39 x 6.02 kJ = 8.36 kJ

qtotal = 0.339 kJ + 8.36 kJ = 8.70 kJ

with these calculations, we can now proceed to answer the question:

(a) False AH is theheat capacity for the melting.

(b) True as we determined above

(c) False we only have one phase change, from solid (ice) to liquid

(d) True as calculated above

(e) False as determined in our calculations

why is arctic sea ice important? arctic sea ice keeps the polar regions cool and moderate global climate. sea ice has a bright surface; 80 percent of the sunlight that strikes it is reflected back into space. as sea ice melts in the summer, it exposes the dark ocean surface.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3