8 g ice at -5 degree c is mixed with 30 g water at 20 degree c, find final temperature of the mixture

To find the final temperature of the mixture when 8 grams of ice at -5°C is mixed with 30 grams of water at 20°C, we need to consider the heat exchange between the ice and the water. The heat lost by the warmer water will be equal to the heat gained by the ice as it warms up and eventually melts. Let's break this down step by step.

Understanding the Components

We have two substances involved in this scenario:

• Ice: 8 g at -5°C
• Water: 30 g at 20°C

Key Concepts

To solve this problem, we will use the principle of conservation of energy, which states that energy lost by the warmer substance (water) will equal the energy gained by the colder substance (ice). The process involves several steps:

• Heating the ice from -5°C to 0°C
• Melting the ice at 0°C to water
• Heating the resulting water from melted ice to the final temperature
• Cooling the warm water from 20°C to the final temperature

Calculating Heat Transfer

Let’s denote:

• m_ice = 8 g
• m_water = 30 g
• c_ice = 2.09 J/g°C (specific heat capacity of ice)
• c_water = 4.18 J/g°C (specific heat capacity of water)
• L_f = 334 J/g (latent heat of fusion for ice)

Step 1: Heating the Ice

The first step is to heat the ice from -5°C to 0°C. The heat required (Q_ice) can be calculated using the formula:

Q_ice = m_ice × c_ice × ΔT

Where ΔT is the change in temperature (0 - (-5) = 5°C).

So,

Q_ice = 8 g × 2.09 J/g°C × 5°C = 83.6 J

Step 2: Melting the Ice

Next, we need to melt the ice at 0°C. The heat required for this (Q_melt) is:

Q_melt = m_ice × L_f

Thus,

Q_melt = 8 g × 334 J/g = 2672 J

Step 3: Total Heat Gained by Ice

The total heat gained by the ice (Q_{total, ice}) is the sum of the heat to warm the ice and the heat to melt it:

Q_{total, ice} = Q_ice + Q_melt = 83.6 J + 2672 J = 2755.6 J

Step 4: Cooling the Water

Now, let's calculate how much heat the warm water can lose as it cools down to the final temperature (T_final). The heat lost by the water (Q_water) can be expressed as:

Q_water = m_water × c_water × (T_initial - T_final)

Substituting the values:

Q_water = 30 g × 4.18 J/g°C × (20°C - T_final)

Q_water = 125.4 J/°C × (20 - T_final)

Setting Up the Equation

Since the heat lost by the water equals the heat gained by the ice, we can set up the equation:

125.4 J/°C × (20 - T_final) = 2755.6 J

Solving for T_final

Now, we can solve for T_final:

125.4 × (20 - T_final) = 2755.6

2508 - 125.4T_final = 2755.6

-125.4T_final = 2755.6 - 2508

-125.4T_final = 247.6

T_final = -247.6 / -125.4 ≈ 1.97°C

Final Result

The final temperature of the mixture, after the ice has melted and reached thermal equilibrium with the water, is approximately 1.97°C. This demonstrates how heat transfer works between substances at different temperatures, leading to a balanced final state.