Question #69008

10 g of ice at -10degree Celsius is added to 10g of water at 85 degree Celsius. What is the final temperature and amount of ice left in the system ( system is kept inside an ideal insulator

Expert's answer

First, let us assume that all added ice will be melted. Thus, the heat balance will follow the equation:

m_i c_i ∆t_1+δ_i m_i+m_i c_w ∆t_2=m_w c_w ∆t_3

m_i-mass of ice,m_w-mass of water,c_i-heat capacity of ice(2100 J/(kg∙K)),c_w-heat capacity of water(4184 J/(kg∙K)),δ_i-melting heat capacity of ice(330 kJ/kg),∆t_1-temperature change upon the heating ice from-10℃ to 0℃,∆t_2-temperature change upon the heating melted ice from 0℃ to an unknown temperature and ∆t_3- temperature change upon the cooling hot water from 85℃ to an unknown temperature.

Water and ice possess the same mass and can be removed from the equation:

c_i ∆t_1+δ_i+c_w ∆t_2=c_w ∆t_3

,where ∆t_1=10℃,∆t_2=t_x-t_0 and ∆t_3=t_85-t_x

∆t_3-∆t_2=t_85-t_x-t_x+t_0=t_85-2t_x

thus,t_x=(t_85∙c_w-c_i∙∆t_1-δ_i)/(2c_w )=0.55℃

The final temperature of the mixture will be 0.55 °C and this is due to the ice completely melted (all 10g).

m_i c_i ∆t_1+δ_i m_i+m_i c_w ∆t_2=m_w c_w ∆t_3

m_i-mass of ice,m_w-mass of water,c_i-heat capacity of ice(2100 J/(kg∙K)),c_w-heat capacity of water(4184 J/(kg∙K)),δ_i-melting heat capacity of ice(330 kJ/kg),∆t_1-temperature change upon the heating ice from-10℃ to 0℃,∆t_2-temperature change upon the heating melted ice from 0℃ to an unknown temperature and ∆t_3- temperature change upon the cooling hot water from 85℃ to an unknown temperature.

Water and ice possess the same mass and can be removed from the equation:

c_i ∆t_1+δ_i+c_w ∆t_2=c_w ∆t_3

,where ∆t_1=10℃,∆t_2=t_x-t_0 and ∆t_3=t_85-t_x

∆t_3-∆t_2=t_85-t_x-t_x+t_0=t_85-2t_x

thus,t_x=(t_85∙c_w-c_i∙∆t_1-δ_i)/(2c_w )=0.55℃

The final temperature of the mixture will be 0.55 °C and this is due to the ice completely melted (all 10g).

Learn more about our help with Assignments: Inorganic Chemistry

## Comments

## Leave a comment