10 gm ice at –10°C, 10 gm water at 20°C and 2g steam at 100°C are mixed with each other then final equilibrium temperature.

Answer

To solve the problem of finding the final equilibrium temperature when mixing 10 g of ice at -10°C, 10 g of water at 20°C, and 2 g of steam at 100°C, we will apply the principle of calorimetry. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the heat required to melt the ice and raise its temperature to 0°C
1. **Heating the ice from -10°C to 0°C:**
- Mass of ice (m_ice) = 10 g
- Specific heat of ice (c_ice) = 0.5 cal/g°C
- Temperature change (ΔT) = 0 - (-10) = 10°C
- Heat required (Q1) = m_ice * c_ice * ΔT
- Q1 = 10 g * 0.5 cal/g°C * 10°C = 50 calories

2. **Melting the ice at 0°C:**
- Latent heat of fusion (L_f) = 80 cal/g
- Heat required to melt the ice (Q2) = m_ice * L_f
- Q2 = 10 g * 80 cal/g = 800 calories

3. **Total heat required to convert ice at -10°C to water at 0°C:**
- Total heat (Q_total_ice) = Q1 + Q2
- Q_total_ice = 50 calories + 800 calories = 850 calories

### Step 2: Calculate the heat released by the steam when it condenses
1. **Condensation of steam at 100°C:**
- Mass of steam (m_steam) = 2 g
- Latent heat of vaporization (L_v) = 540 cal/g
- Heat released by steam (Q_steam) = m_steam * L_v
- Q_steam = 2 g * 540 cal/g = 1080 calories

### Step 3: Determine if the heat from the steam is sufficient to melt the ice
- Since Q_steam (1080 calories) > Q_total_ice (850 calories), all the ice will melt, and there will be excess heat available.

### Step 4: Calculate the excess heat after melting the ice
- Excess heat = Q_steam - Q_total_ice
- Excess heat = 1080 calories - 850 calories = 230 calories

### Step 5: Set up the equation for the final equilibrium temperature
1. **Components after mixing:**
- Water from melted ice = 10 g at 0°C
- Original water = 10 g at 20°C
- Water from condensed steam = 2 g at 100°C

2. **Let T be the final equilibrium temperature.**
- Heat absorbed by the melted ice (10 g at 0°C):
- Q_absorbed_ice = 10 g * 1 cal/g°C * (T - 0) = 10T calories
- Heat absorbed by the original water (10 g at 20°C):
- Q_absorbed_water = 10 g * 1 cal/g°C * (T - 20) = 10(T - 20) = 10T - 200 calories
- Heat absorbed by the condensed steam (2 g at 100°C):
- Q_absorbed_steam = 2 g * 1 cal/g°C * (T - 100) = 2(T - 100) = 2T - 200 calories

3. **Total heat absorbed:**
- Total heat absorbed = Q_absorbed_ice + Q_absorbed_water + Q_absorbed_steam
- Total heat absorbed = 10T + (10T - 200) + (2T - 200)
- Total heat absorbed = 22T - 400 calories

### Step 6: Set the heat absorbed equal to the excess heat
- Set the total heat absorbed equal to the excess heat:
- 22T - 400 = 230

### Step 7: Solve for T
1. Rearranging the equation:
- 22T = 230 + 400
- 22T = 630
- T = 630 / 22
- T = 28.6363°C

### Final Answer
The final equilibrium temperature is approximately **28.64°C**.

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