Two 20.0-g ice cubes at –15.0 °c are placed into 265 g of water at 25.0 °c. assuming no energy is transferred to or from the surroundings, calculate the final temperature, tf, of the water after all the ice melts. heat capacity of h20(s) is 37.7j/kmol. heat capacity of h20(l) is 75.3j/kmol. enthalpy of fusion of h20 is 6.01j/kmol.

10.3 °C

Step-by-step explanation:

This is an isolated system so

heat gained by ice + heat lost by water = 0

Heat to warm ice + heat to melt ice + heat to warm melt + heat lost by water = 0

     q₁     +     q₂        +      q₃      +     q₄      = 0

n₁C₁ΔT₁ + n₁ΔH_fus + n₁C₃ΔT₃ + n₄C₃ΔT₄  = 0

Step 1: Calculate q₁

n₁ = 40.0 g × 1/18.02

n₁ = 2.220 mol

C₁ = 37.7 J·K⁻¹mol⁻¹

ΔT₁ = T_f – T_i  

ΔT₁ = 0.0 – (-15)

ΔT₁ = 15 °C = 15 K

q₁ = 2.220 × 37.7 × 15  

q₁ = 1255 J

q₁ = 1.255 kJ

Step 2. Calculate q₂

ΔH_fus = 6.01 kJ·mol⁻¹

q₂ = 2.220 × 6.01

q₂ = 13.34 kJ

Step 3: Calculate q₃

C₃ = 75.3 J·K⁻¹mol⁻¹

ΔT₃ = T_f – T_i  = x – 0 °C  

ΔT₃ = x °C = x K

q₃ = 2.220 × 75.3 × x

q₃ = 167x J

q₃ = 0.167x kJ

Step 4. Calculate q₄

n₄ = 265 × 1/18.02

n₄ = 14.71 mol

ΔT₄ = T_f – T_i  = x – 25.0 °C  

q₄ = 14.71 × 75.3 × (x – 25.0)

q₄ = 1107 × (x – 25.0)

q₄ = 1107x  – 27 680 J

q₄ = 1.107x – 27.68 kJ

Step 5. Solve for x

1.255 + 13.34 + 0.167x + 1.107x - 27.68 = 0

1.274x - 13.08 = 0

1.274x = 13.08

x = 13.08/1.274

x = 10.3 °C

The final temperature is -10.3 °C.