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QuestionA 5.0-gram ice cube at - 2.0 degrees Celsius is dropped into 60 grams of liquid water at 12 degrees Celsius. If the ice and water are isolated from the environment, what is the final temperature of the combined liquid water when all of the ice has melted and the mixture reaches thermal equilibrium? □\square A \quad degrees Celsius □\square Hide hint for Question 1 Report your result to 2 significant figures.

Studdy Solution

STEP 1

1. The system is isolated, meaning no heat is lost to the environment.
2. The specific heat capacity of ice is cice=2.09 J/g°C c_{\text{ice}} = 2.09 \, \text{J/g°C} .
3. The specific heat capacity of water is cwater=4.18 J/g°C c_{\text{water}} = 4.18 \, \text{J/g°C} .
4. The heat of fusion for ice is Lf=334 J/g L_f = 334 \, \text{J/g} .
5. We need to find the final temperature of the water after the ice has melted and thermal equilibrium is reached.

STEP 2

1. Calculate the heat required to raise the temperature of the ice to 0°C.
2. Calculate the heat required to melt the ice.
3. Calculate the heat lost by the water as it cools to the final temperature.
4. Set up an equation for thermal equilibrium.
5. Solve for the final temperature.

STEP 3

Calculate the heat required to raise the temperature of the ice from -2.0°C to 0°C.
Use the formula: q1=mice⋅cice⋅ΔT q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T
Where: - mice=5.0 g m_{\text{ice}} = 5.0 \, \text{g} - cice=2.09 J/g°C c_{\text{ice}} = 2.09 \, \text{J/g°C} - ΔT=0−(−2.0)=2.0 °C \Delta T = 0 - (-2.0) = 2.0 \, \text{°C}
q1=5.0 g⋅2.09 J/g°C⋅2.0 °C=20.9 J q_1 = 5.0 \, \text{g} \cdot 2.09 \, \text{J/g°C} \cdot 2.0 \, \text{°C} = 20.9 \, \text{J}

STEP 4

Calculate the heat required to melt the ice.
Use the formula: q2=mice⋅Lf q_2 = m_{\text{ice}} \cdot L_f
Where: - Lf=334 J/g L_f = 334 \, \text{J/g}
q2=5.0 g⋅334 J/g=1670 J q_2 = 5.0 \, \text{g} \cdot 334 \, \text{J/g} = 1670 \, \text{J}

STEP 5

Calculate the heat lost by the water as it cools to the final temperature Tf T_f .
Use the formula: q3=mwater⋅cwater⋅(Tinitial water−Tf) q_3 = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{initial water}} - T_f)
Where: - mwater=60 g m_{\text{water}} = 60 \, \text{g} - cwater=4.18 J/g°C c_{\text{water}} = 4.18 \, \text{J/g°C} - Tinitial water=12 °C T_{\text{initial water}} = 12 \, \text{°C}
q3=60 g⋅4.18 J/g°C⋅(12−Tf) q_3 = 60 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot (12 - T_f)

STEP 6

Set up an equation for thermal equilibrium.
The heat gained by the ice must equal the heat lost by the water: q1+q2=q3 q_1 + q_2 = q_3
Substitute the known values: 20.9 J+1670 J=60 g⋅4.18 J/g°C⋅(12−Tf) 20.9 \, \text{J} + 1670 \, \text{J} = 60 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot (12 - T_f)

STEP 7

Solve for the final temperature Tf T_f .
1690.9=60⋅4.18⋅(12−Tf) 1690.9 = 60 \cdot 4.18 \cdot (12 - T_f)
1690.9=250.8⋅(12−Tf) 1690.9 = 250.8 \cdot (12 - T_f)
Divide both sides by 250.8: 12−Tf=1690.9250.8 12 - T_f = \frac{1690.9}{250.8}
12−Tf=6.74 12 - T_f = 6.74
Tf=12−6.74 T_f = 12 - 6.74
Tf=5.26 T_f = 5.26
Report the result to 2 significant figures: Tf=5.3 °C T_f = 5.3 \, \text{°C}
The final temperature of the combined liquid water is:
5.3 °C \boxed{5.3 \, \text{°C}}

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