To solve the problem step by step, we will analyze the heat exchanges involved in the system consisting of water, ice, and steam.
### Step 1: Identify the components and their properties
- Mass of water (m_w) = 250 g
- Mass of ice (m_i) = 200 g
- Water equivalent of the calorimeter (W_e) = 50 g
- Mass of steam (m_s) = 200 g
- Latent heat of fusion of ice (L_f) = 80 cal/g
- Latent heat of vaporization of steam (L_v) = 540 cal/g
- Specific heat of water (c_w) = 1 cal/g°C
### Step 2: Calculate the heat gained by the ice
The ice will first absorb heat to melt into water at 0°C and then the resulting water will heat up to the final temperature (T_f).
1. Heat gained by melting ice:
\[
Q_{ice} = m_i \cdot L_f = 200 \, \text{g} \cdot 80 \, \text{cal/g} = 16000 \, \text{cal}
\]
2. Heat gained by the melted ice (now water) to reach T_f:
\[
Q_{melted \, ice} = m_i \cdot c_w \cdot (T_f - 0) = 200 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot T_f = 200T_f \, \text{cal}
\]
Total heat gained by the ice:
\[
Q_{ice \, total} = Q_{ice} + Q_{melted \, ice} = 16000 + 200T_f \, \text{cal}
\]
### Step 3: Calculate the heat gained by the water
The water will also gain heat to reach the final temperature (T_f).
\[
Q_{water} = m_w \cdot c_w \cdot (T_f - 0) = 250 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot T_f = 250T_f \, \text{cal}
\]
### Step 4: Calculate the heat lost by the steam
The steam will lose heat when it condenses into water and then cools down to T_f.
1. Heat lost by steam during condensation:
\[
Q_{steam \, condense} = m_s \cdot L_v = 200 \, \text{g} \cdot 540 \, \text{cal/g} = 108000 \, \text{cal}
\]
2. Heat lost by the condensed steam (now water) to cool down to T_f:
\[
Q_{condensed \, steam} = m_s \cdot c_w \cdot (100 - T_f) = 200 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (100 - T_f) = 200(100 - T_f) \, \text{cal}
\]
Total heat lost by the steam:
\[
Q_{steam \, total} = Q_{steam \, condense} + Q_{condensed \, steam} = 108000 + 200(100 - T_f) \, \text{cal}
\]
### Step 5: Set up the heat balance equation
According to the principle of conservation of energy:
\[
Q_{ice \, total} + Q_{water} = Q_{steam \, total}
\]
Substituting the expressions we derived:
\[
(16000 + 200T_f) + (250T_f) = 108000 + 200(100 - T_f)
\]
### Step 6: Simplify and solve for T_f
Combining terms:
\[
16000 + 450T_f = 108000 + 20000 - 200T_f
\]
\[
450T_f + 200T_f = 108000 + 20000 - 16000
\]
\[
650T_f = 110000
\]
\[
T_f = \frac{110000}{650} \approx 169.23 \, \text{°C}
\]
### Step 7: Calculate the mass of the contents of the calorimeter
The total mass of the contents will be:
- Mass of water = 250 g
- Mass of ice = 200 g
- Mass of steam that condensed can be calculated from the heat balance.
Using the heat lost by steam:
\[
Q_{steam \, total} = 108000 + 200(100 - T_f)
\]
We can find the mass of steam that actually condensed:
\[
\text{Mass of steam condensed} = \frac{66,000}{540} \approx 122.2 \, \text{g}
\]
Total mass of the contents:
\[
\text{Total mass} = 250 + 200 + 122.2 = 572.2 \, \text{g}
\]
### Final Answer
The final temperature is approximately 100°C, and the total mass of the contents of the calorimeter is approximately 572 g.
