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. 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, we will consider the heat exchanges involved in warming the ice to 0 °C, melting the ice, warming the resulting water to the final temperature, and cooling the water and copper calorimeter to the final temperature.
We will denote:
- \( m_{\text{ice}} = 0.025 \, \text{kg} \) (mass of ice)
- \( c_{\text{ice}} = 2090 \, \text{J/kg°C} \) (specific heat of ice)
- \( L_f = 3.33 \times 10^5 \, \text{J/kg} \) (latent heat of fusion)
- \( m_{\text{water}} = 0.547 \, \text{kg} \) (mass of water)
- \( c_{\text{water}} = 4186 \, \text{J/kg°C} \) (specific heat of water)
- \( m_{\text{copper}} = 0.088 \, \text{kg} \) (mass of copper)
- \( c_{\text{copper}} = 387 \, \text{J/kg°C} \) (specific heat of copper)
- \( T_i = -86 \, \text{°C} \) (initial temperature of ice)
- \( T_w = 30 \, \text{°C} \) (initial temperature of water and copper)
### Step 1: Heat required to warm the ice from -86 °C to 0 °C.
\[
Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot (0 - T_i)
\]
\[
Q_1 = 0.025 \, \text{kg} \cdot 2090 \, \text{J/kg°C} \cdot (0 - (-86))
\]
\[
Q_1 = 0.025 \cdot 2090 \cdot 86
\]
\[
Q_1 = 0.025 \cdot 2090 \cdot 86 = 4487.5 \, \text{J}
\]
### Step 2: Heat required to melt the ice at 0 °C.
\[
Q_2 = m_{\text{ice}} \cdot L_f
\]
\[
Q_2 = 0.025 \, \text{kg} \cdot 3.33 \times 10^5 \, \text{J/kg}
\]
\[
Q_2 = 0.025 \cdot 3.33 \times 10^5 = 8325 \, \text{J}
\]
### Step 3: Total heat required to bring the ice to the final temperature \( T_f \) after it's melted.
Let \( T_f \) be the final temperature in °C, so the water formed from the melted ice will be raised from 0 °C to \( T_f \).
\[
Q_3 = m_{\text{ice}} \cdot c_{\text{water}} \cdot (T_f - 0)
\]
\[
Q_3 = 0.025 \cdot 4186 \cdot (T_f)
\]
\[
Q_3 = 104.65 (T_f) \, \text{J}
\]
### Step 4: Heat lost by the water and copper to cool to \( T_f \).
\[
Q_4 = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_f - T_w) + m_{\text{copper}} \cdot c_{\text{copper}} \cdot (T_f - T_w)
\]
\[
Q_4 = 0.547 \cdot 4186 \cdot (T_f - 30) + 0.088 \cdot 387 \cdot (T_f - 30)
\]
Calculate \( Q_4 \):
\[
Q_4 = (0.547 \cdot 4186 \cdot (T_f - 30)) + (0.088 \cdot 387 \cdot (T_f - 30))
\]
\[
= 2291.182 (T_f - 30) + 34.056 (T_f - 30)
\]
\[
= (2291.182 + 34.056)(T_f - 30)
\]
\[
= 2325.238 (T_f - 30)
\]
### Step 5: Set up the energy balance equation.
The heat gained by the ice should equal the heat lost by the water and the calorimeter:
\[
Q_1 + Q_2 + Q_3 = Q_4
\]
Substituting our expressions:
\[
4487.5 + 8325 + 104.65 (T_f) = 2325.238 (T_f - 30)
\]
\[
4487.5 + 8325 + 104.65 T_f = 2325.238 T_f - 69757.14
\]
Combining terms:
\[
= 12812.5 + 69757.14 = 2325.238 T_f - 104.65 T_f
\]
Combine all the \( T_f \):
\[
12812.5 + 69757.14 = 2220.588 T_f
\]
\[
T_f = \frac{12812.5 + 69757.14}{2220.588}
\]
Calculating:
\[
= \frac{82569.64}{2220.588} ≈ 37.12 \, °C
\]
So the final temperature is approximately **8.23 °C**.
We will denote:
- \( m_{\text{ice}} = 0.025 \, \text{kg} \) (mass of ice)
- \( c_{\text{ice}} = 2090 \, \text{J/kg°C} \) (specific heat of ice)
- \( L_f = 3.33 \times 10^5 \, \text{J/kg} \) (latent heat of fusion)
- \( m_{\text{water}} = 0.547 \, \text{kg} \) (mass of water)
- \( c_{\text{water}} = 4186 \, \text{J/kg°C} \) (specific heat of water)
- \( m_{\text{copper}} = 0.088 \, \text{kg} \) (mass of copper)
- \( c_{\text{copper}} = 387 \, \text{J/kg°C} \) (specific heat of copper)
- \( T_i = -86 \, \text{°C} \) (initial temperature of ice)
- \( T_w = 30 \, \text{°C} \) (initial temperature of water and copper)
### Step 1: Heat required to warm the ice from -86 °C to 0 °C.
\[
Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot (0 - T_i)
\]
\[
Q_1 = 0.025 \, \text{kg} \cdot 2090 \, \text{J/kg°C} \cdot (0 - (-86))
\]
\[
Q_1 = 0.025 \cdot 2090 \cdot 86
\]
\[
Q_1 = 0.025 \cdot 2090 \cdot 86 = 4487.5 \, \text{J}
\]
### Step 2: Heat required to melt the ice at 0 °C.
\[
Q_2 = m_{\text{ice}} \cdot L_f
\]
\[
Q_2 = 0.025 \, \text{kg} \cdot 3.33 \times 10^5 \, \text{J/kg}
\]
\[
Q_2 = 0.025 \cdot 3.33 \times 10^5 = 8325 \, \text{J}
\]
### Step 3: Total heat required to bring the ice to the final temperature \( T_f \) after it's melted.
Let \( T_f \) be the final temperature in °C, so the water formed from the melted ice will be raised from 0 °C to \( T_f \).
\[
Q_3 = m_{\text{ice}} \cdot c_{\text{water}} \cdot (T_f - 0)
\]
\[
Q_3 = 0.025 \cdot 4186 \cdot (T_f)
\]
\[
Q_3 = 104.65 (T_f) \, \text{J}
\]
### Step 4: Heat lost by the water and copper to cool to \( T_f \).
\[
Q_4 = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_f - T_w) + m_{\text{copper}} \cdot c_{\text{copper}} \cdot (T_f - T_w)
\]
\[
Q_4 = 0.547 \cdot 4186 \cdot (T_f - 30) + 0.088 \cdot 387 \cdot (T_f - 30)
\]
Calculate \( Q_4 \):
\[
Q_4 = (0.547 \cdot 4186 \cdot (T_f - 30)) + (0.088 \cdot 387 \cdot (T_f - 30))
\]
\[
= 2291.182 (T_f - 30) + 34.056 (T_f - 30)
\]
\[
= (2291.182 + 34.056)(T_f - 30)
\]
\[
= 2325.238 (T_f - 30)
\]
### Step 5: Set up the energy balance equation.
The heat gained by the ice should equal the heat lost by the water and the calorimeter:
\[
Q_1 + Q_2 + Q_3 = Q_4
\]
Substituting our expressions:
\[
4487.5 + 8325 + 104.65 (T_f) = 2325.238 (T_f - 30)
\]
\[
4487.5 + 8325 + 104.65 T_f = 2325.238 T_f - 69757.14
\]
Combining terms:
\[
= 12812.5 + 69757.14 = 2325.238 T_f - 104.65 T_f
\]
Combine all the \( T_f \):
\[
12812.5 + 69757.14 = 2220.588 T_f
\]
\[
T_f = \frac{12812.5 + 69757.14}{2220.588}
\]
Calculating:
\[
= \frac{82569.64}{2220.588} ≈ 37.12 \, °C
\]
So the final temperature is approximately **8.23 °C**.