Question
A 25 g block of ice is cooled to −86 ◦C. It is added to 547 g of water in an 88 g copper calorimeter at a temperature of 30◦C. Find the final temperature of the mixture. The specific heat of copper is 387 J/kg ·◦C and of ice is 2090 J/kg ·◦C. The latent heat of fusion of water is 3.33 × 105 J/kg and its specific heat is 4186 J/kg ·◦C. Answer in units of ◦C.
Answers
GPT-4o mini
To find the final temperature of the system, we'll use the principle of conservation of energy. The heat lost by the water and the copper calorimeter will equal the heat gained by the ice.
Given:
- Mass of ice (\( m_{ice} \)): 25 g = 0.025 kg
- Initial temperature of ice (\( T_{ice} \)): -86 °C
- Mass of water (\( m_{water} \)): 547 g = 0.547 kg
- Initial temperature of water (\( T_{water} \)): 30 °C
- Mass of copper calorimeter (\( m_{copper} \)): 88 g = 0.088 kg
- Initial temperature of copper (\( T_{copper} \)): 30 °C
- Specific heat of ice (\( c_{ice} \)): 2090 J/kg·°C
- Latent heat of fusion of ice (\( L_f \)): \( 3.33 \times 10^5 \, \text{J/kg} \)
- Specific heat of water (\( c_{water} \)): 4186 J/kg·°C
- Specific heat of copper (\( c_{copper} \)): 387 J/kg·°C
We will consider two scenarios:
1. The ice first warms up to 0 °C and then melts.
2. The resulting water from melted ice is then warmed to the final temperature \( T_f \) along with the original water and the calorimeter.
### Step 1: Heat gained by the ice
1. Heating the ice from -86 °C to 0 °C:
\[
Q_1 = m_{ice} \cdot c_{ice} \cdot (0 - T_{ice}) = 0.025 \, \text{kg} \cdot 2090 \, \text{J/kg·°C} \cdot (0 - (-86)) = 0.025 \cdot 2090 \cdot 86
\]
Calculating \( Q_1 \):
\[
Q_1 = 0.025 \cdot 2090 \cdot 86 = 4480.75 \, \text{J}
\]
2. Melting the ice at 0 °C:
\[
Q_2 = m_{ice} \cdot L_f = 0.025 \, \text{kg} \cdot 3.33 \times 10^5 \, \text{J/kg} = 8335 \, \text{J}
\]
3. Heating the resulting water to \( T_f \):
\[
Q_3 = m_{ice} \cdot c_{water} \cdot (T_f - 0) = 0.025 \cdot 4186 \cdot T_f = 104.65 T_f
\]
### Total heat gained by ice:
\[
Q_{gained} = Q_1 + Q_2 + Q_3 = 4480.75 + 8335 + 104.65T_f
\]
\[
Q_{gained} = 12815.75 + 104.65T_f
\]
### Step 2: Heat lost by water and copper
1. Water cooling from 30 °C to \( T_f \):
\[
Q_{lost, water} = m_{water} \cdot c_{water} \cdot (T_{water} - T_f) = 0.547 \, \text{kg} \cdot 4186 \, \text{J/kg·°C} \cdot (30 - T_f) = 22873.62(30 - T_f)
\]
2. Copper cooling from 30 °C to \( T_f \):
\[
Q_{lost, copper} = m_{copper} \cdot c_{copper} \cdot (T_{copper} - T_f) = 0.088 \, \text{kg} \cdot 387 \, \text{J/kg·°C} \cdot (30 - T_f) = 340.68(30 - T_f)
\]
### Total heat lost:
\[
Q_{lost} = Q_{lost, water} + Q_{lost, copper} = 22873.62(30 - T_f) + 340.68(30 - T_f)
\]
\[
Q_{lost} = (22873.62 + 340.68)(30 - T_f) = 23214.3(30 - T_f)
\]
### Step 3: Setting heat gain equal to heat loss
\[
12815.75 + 104.65T_f = 23214.3(30 - T_f)
\]
### Step 4: Expand and solve for \( T_f \)
Expanding:
\[
12815.75 + 104.65T_f = 696429 - 23214.3T_f
\]
Rearranging gives:
\[
104.65T_f + 23214.3T_f = 696429 - 12815.75
\]
\[
23318.95T_f = 683613.25
\]
\[
T_f = \frac{683613.25}{23318.95} \approx 29.4 \, \text{°C}
\]
### Conclusion:
The final temperature of the mixture is approximately \(\boxed{29.4 \, \text{°C}}\).
Given:
- Mass of ice (\( m_{ice} \)): 25 g = 0.025 kg
- Initial temperature of ice (\( T_{ice} \)): -86 °C
- Mass of water (\( m_{water} \)): 547 g = 0.547 kg
- Initial temperature of water (\( T_{water} \)): 30 °C
- Mass of copper calorimeter (\( m_{copper} \)): 88 g = 0.088 kg
- Initial temperature of copper (\( T_{copper} \)): 30 °C
- Specific heat of ice (\( c_{ice} \)): 2090 J/kg·°C
- Latent heat of fusion of ice (\( L_f \)): \( 3.33 \times 10^5 \, \text{J/kg} \)
- Specific heat of water (\( c_{water} \)): 4186 J/kg·°C
- Specific heat of copper (\( c_{copper} \)): 387 J/kg·°C
We will consider two scenarios:
1. The ice first warms up to 0 °C and then melts.
2. The resulting water from melted ice is then warmed to the final temperature \( T_f \) along with the original water and the calorimeter.
### Step 1: Heat gained by the ice
1. Heating the ice from -86 °C to 0 °C:
\[
Q_1 = m_{ice} \cdot c_{ice} \cdot (0 - T_{ice}) = 0.025 \, \text{kg} \cdot 2090 \, \text{J/kg·°C} \cdot (0 - (-86)) = 0.025 \cdot 2090 \cdot 86
\]
Calculating \( Q_1 \):
\[
Q_1 = 0.025 \cdot 2090 \cdot 86 = 4480.75 \, \text{J}
\]
2. Melting the ice at 0 °C:
\[
Q_2 = m_{ice} \cdot L_f = 0.025 \, \text{kg} \cdot 3.33 \times 10^5 \, \text{J/kg} = 8335 \, \text{J}
\]
3. Heating the resulting water to \( T_f \):
\[
Q_3 = m_{ice} \cdot c_{water} \cdot (T_f - 0) = 0.025 \cdot 4186 \cdot T_f = 104.65 T_f
\]
### Total heat gained by ice:
\[
Q_{gained} = Q_1 + Q_2 + Q_3 = 4480.75 + 8335 + 104.65T_f
\]
\[
Q_{gained} = 12815.75 + 104.65T_f
\]
### Step 2: Heat lost by water and copper
1. Water cooling from 30 °C to \( T_f \):
\[
Q_{lost, water} = m_{water} \cdot c_{water} \cdot (T_{water} - T_f) = 0.547 \, \text{kg} \cdot 4186 \, \text{J/kg·°C} \cdot (30 - T_f) = 22873.62(30 - T_f)
\]
2. Copper cooling from 30 °C to \( T_f \):
\[
Q_{lost, copper} = m_{copper} \cdot c_{copper} \cdot (T_{copper} - T_f) = 0.088 \, \text{kg} \cdot 387 \, \text{J/kg·°C} \cdot (30 - T_f) = 340.68(30 - T_f)
\]
### Total heat lost:
\[
Q_{lost} = Q_{lost, water} + Q_{lost, copper} = 22873.62(30 - T_f) + 340.68(30 - T_f)
\]
\[
Q_{lost} = (22873.62 + 340.68)(30 - T_f) = 23214.3(30 - T_f)
\]
### Step 3: Setting heat gain equal to heat loss
\[
12815.75 + 104.65T_f = 23214.3(30 - T_f)
\]
### Step 4: Expand and solve for \( T_f \)
Expanding:
\[
12815.75 + 104.65T_f = 696429 - 23214.3T_f
\]
Rearranging gives:
\[
104.65T_f + 23214.3T_f = 696429 - 12815.75
\]
\[
23318.95T_f = 683613.25
\]
\[
T_f = \frac{683613.25}{23318.95} \approx 29.4 \, \text{°C}
\]
### Conclusion:
The final temperature of the mixture is approximately \(\boxed{29.4 \, \text{°C}}\).