Multiple Choice
If a hot metal with a mass of 5.21 grams undergoes a temperature decrease of 18.18 °C and releases 150 J of heat, what is the specific heat capacity of the metal?
8. Thermochemistry
Heat Capacity
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The heat capacity of a calorimeter was determined by adding 50.0 g of water at 45.0°C to the coffee cup calorimeter that contained 50.0 g of water at 20.0°C. The final temperature was measured to be 31.8°C. Calculate the heat capacity of the calorimeter (C_cal) in J/°C.
A
25.0 J/°C
B
20.3 J/°C
C
10.5 J/°C
D
15.8 J/°C
Verified step by step guidance1
Identify the masses and temperatures of the water samples: 50.0 g of water at 45.0°C and 50.0 g of water at 20.0°C. The final temperature is 31.8°C.
Calculate the heat lost by the hot water using the formula: \( q_{\text{hot}} = m_{\text{hot}} \cdot c \cdot (T_{\text{final}} - T_{\text{initial, hot}}) \), where \( m_{\text{hot}} \) is the mass of the hot water, \( c \) is the specific heat capacity of water (4.18 J/g°C), and \( T \) represents temperature.
Calculate the heat gained by the cold water using the formula: \( q_{\text{cold}} = m_{\text{cold}} \cdot c \cdot (T_{\text{final}} - T_{\text{initial, cold}}) \), where \( m_{\text{cold}} \) is the mass of the cold water.
Apply the principle of conservation of energy: the heat lost by the hot water is equal to the heat gained by the cold water plus the heat absorbed by the calorimeter. Set up the equation: \( q_{\text{hot}} = q_{\text{cold}} + C_{\text{cal}} \cdot (T_{\text{final}} - T_{\text{initial, cold}}) \).
Rearrange the equation to solve for the heat capacity of the calorimeter \( C_{\text{cal}} \): \( C_{\text{cal}} = \frac{q_{\text{hot}} - q_{\text{cold}}}{T_{\text{final}} - T_{\text{initial, cold}}} \).
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