Question

Two 20.0-g ice cubes at –18.0 °C are placed into 245 g of water at 25.0 °C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melts. Please show work.

Heat capacity of H20(s): 37.7 J/(mol x K)

Heat capacity of H20(l): 75.3 J/(mol x K)

Enthalpy of fusion of H20: 6.01 kJ/mol

Answer 1

Mass of ice =2* 20.0 g =40.0 g

moles of ice =

moles of ice =m_ice=2.22 mol (since heat capacities are given in J/mol. K, we are converting grams to moles and deg C to K)

Ice cube temperature , T₁= -18 deg C = 255 K

first we require to heat ice to 0 deg C = 273 K

Heat capacity of ice =C_ice = 37.7 J/(mol. K)

Amount heat required to rise the temperature from -18 to 0 ^oC =

q₁ = m_iceC_ice(T₂-T₁)

Next

enthalpy of fusion =

Amount of heat required to melt ice =

After ice melts to water, heat will be absorbed to rise the temperature

Let final temperature be = T₂ = ?k

Heat absorbed =

Total amount of heat absorbed by ice = q₁+q₂+q₃ = amount of heat given by water

Amount of heat given by water =

mass of water = 245 g

moles of water = m_l =

Heat capacity of H₂0_(l)= 75.3 J/(mol x K)

Initial temperature =T₁ = 298 K

Final temperature =T2 =? K

T2 = 282 K

Final temperature = T2 = 282 K = 9 ^oC