Answer to Question #260713 in Chemistry for ASDAD

Question

A 20 g sample of ice at -10 °C is mixed with 100 g water at 80 °C. Calculate the

final temperature of the mixture.

Answer 1

The specific heat capacity of ice (C_ice) is 2.108 J g⁻¹ °C⁻¹

The specific heat capacity of water (C_water) is 4.187 J g⁻¹ °C⁻¹

The heat of fusion of ice (ΔH_f) is 333.55 J g⁻¹

T₁ = -10°C

T₂ = 0°C

T₃ = 80°C

Solution:

This problem can be summarized thusly:

1) Ice is heated from -10°C to 0°C.

The heat is calculated by using the specific heat of ice (C_ice) and the equation: q = m × C × ΔT

q₁ = m_ice × C_ice × (T₂ - T₁)

q₁ = (20 g) × (2.108 J g⁻¹ °C⁻¹) × (0 - (-10))°C = 421.6 J

q₁ = 421.6 Joules

2) Ice is melted at 0°C.

The heat is calculated by multiplying the mass of ice by the heat of fusion of ice (ΔH_f).

q₂ = ΔH_f × m_ice = (333.55 J g⁻¹) × (20 g) = 6671 J

q₂ = 6671 Joules

3) 20 g of water at 0°C is heated to the final temperature of the mixture.

The heat is calculated by using the specific heat of water (C_water) and the equation: q = m × C × ΔT

q₃ = m_ice × C_water × (T_f - T₂)

q₃ = (20 g) × (4.187 J g⁻¹ °C⁻¹) × (T_f - 0)°C = 83.74 J × (T_f - 0)

q₃ = 83.74 × T_f Joules

4) 100 g of water at 80°C is cooled to the final temperature of the mixture.

The heat is calculated by using the specific heat of water (C_water) and the equation: q = m × C × ΔT

q₄ = m_water × C_water × (T_f - T₃)

q₄ = (100 g) × (4.187 J g⁻¹ °C⁻¹) × (T_f - 80)°C = 418.7 J × (T_f - 80)

q₄ = 418.7 × T_f - 33496 Joules

Now just sum all of these and the sum must equal to zero; that is, heat gained by ice = heat lost by water.

q₁ + q₂ + q₃ + q₄ = 0

(421.6) + (6671) + (83.74 × T_f) + (418.7 × T_f - 33496) = 0

(-26403.4) + (502.44 × T_f) = 0

502.44 × T_f = 26403.4

T_f = 52.55°C

Answer: The final temperature of the mixture is 52.55°C