Question

To practice Problem-Solving Strategy 12.1 Calorimetry problems.

On a hot summer day, you decide to make some iced tea. First, you brew 1.50 L of hot tea and leave it to steep until it has reached a temperature of Ttea = 75.0 ∘C. You then add 0.975 kg of ice taken from the freezer at a temperature of Tice = 0 ∘C. By the time the mix reaches equilibrium, all of the ice has melted. What is the final temperature Tf of the mixture?

For the purposes of this problem, assume that the tea has the same thermodynamic properties as plain water.

• The specific heat of water is c = 4190 J/kg⋅∘C.
• The heat of fusion of ice is Lf = 3.33×10⁵ J/kg .
• The density of the tea is ρtea = 1.00 kg/L .
• What are the interacting systems in this problem?

Answer 1

Let , T_f be the final temperature of the mixture

m_w = mass of water ( tea ) = 1.5 kg

m_i = mass of ice = 0.975 kg

c_w = specific heat capacity of water

L_f = heat of fusion of water (tea)

Heat lost by water = m_w × c_w × (75 -T_f)

= 1.5 × 4190 × (75 -T_f)

Heat gained by ice = m_i×L_f + m_i ×c_w × T_{​​​​​​f}

= 0.975 × 3.33 × 10⁵ + 0.975 × 4190 × T_f

Assuming no loss of heat to the surrounding

Heat lost by water = heat gained by ice

1.5 × 4190 × (75 - T_f) = 0.975 ×3.33× 10⁵ + 0.975 × 4190 × T_{​​​​​​f}

6285 × (75 - T_{​​​​f}) = 324675 + 4085.25 × T_{​​​​​​f}

471375 - 6285 × T_​​​f = 324675 + 4085.25 × T_{​​​​​​f}

146700 = 10370.25 × T_f

T_f = 14.15 °C

Therefore, final temperature of the mixture T_f = 14.15°C

Interactive systems in this problem are tea and ice as we have assumed no loss of heat to the surroundings.