Respuesta :
Answer:
The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg
Explanation:
Heat gain by ice = Heat lost by water
Thus,
Heat of fusion + [tex]m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)[/tex]
Where, negative sign signifies heat loss
Or,
Heat of fusion + [tex]m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)[/tex]
Heat of fusion = 334 J/g
Heat of fusion of ice with mass x = 334x J/g
For ice:
Mass = x g
Initial temperature = 0 °C
Final temperature = 6 °C
Specific heat of ice = 1.996 J/g°C
For water:
Volume = 353 mL
[tex]Density (\rho)=\frac{Mass(m)}{Volume(V)}[/tex]
Density of water = 1.0 g/mL
So, mass of water = 353 g
Initial temperature = 26 °C
Final temperature = 6 °C
Specific heat of water = 4.186 J/g°C
So,
[tex]334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)[/tex]
[tex]334x+x\times 11.976=29553.16[/tex]
345.976x = 29553.16
x = 85.4197 kg
Thus,
The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg
