- A 40.00 g sample of water at 22.00 C is put in a calorimeter. After 60.00g of water at 65.00C is added to the calorimeter, the mixture is stirred, and the final temperature. is 47.00C. Calculate the heat capacity of the calorimeter in cal/C. Answer 3.2 cal/c. Show how you got this.
- A piece of metal weighing 7.00g is heated to 93.0 C and placed into the came calorimeter described in the previous problem. The calorimeter contains 25.0g of water at 30.0C. If the final temperature of the water is 33.0C, Calculate the specific heat of the heat in cal/g C. answer 0.20 cal/g C. Show how you got this.
The Correct Answer and Explanation is :
Problem 1: Heat Capacity of the Calorimeter
We know that:
- Mass of cool water ((m_1)) = 40.00 g
- Initial temperature of cool water ((T_1)) = 22.00°C
- Mass of warm water ((m_2)) = 60.00 g
- Initial temperature of warm water ((T_2)) = 65.00°C
- Final temperature ((T_f)) = 47.00°C
- Heat capacity of water = 1 cal/g°C
- Heat capacity of calorimeter = (C_{cal}) (unknown)
Step 1: Heat lost by warm water
[
q_{\text{warm water}} = m_2 \cdot c \cdot (T_2 – T_f)
]
[
q_{\text{warm water}} = 60.00 \cdot 1 \cdot (65.00 – 47.00) = 60.00 \cdot 18.00 = 1080 \, \text{cal}
]
Step 2: Heat gained by cool water
[
q_{\text{cool water}} = m_1 \cdot c \cdot (T_f – T_1)
]
[
q_{\text{cool water}} = 40.00 \cdot 1 \cdot (47.00 – 22.00) = 40.00 \cdot 25.00 = 1000 \, \text{cal}
]
Step 3: Heat gained by calorimeter
The heat lost by the warm water is equal to the heat gained by the cool water and the calorimeter:
[
q_{\text{warm water}} = q_{\text{cool water}} + q_{\text{calorimeter}}
]
[
1080 = 1000 + C_{cal} \cdot (T_f – T_1)
]
[
C_{cal} = \frac{1080 – 1000}{47.00 – 22.00} = \frac{80}{25.00} = 3.2 \, \text{cal/°C}
]
Problem 2: Specific Heat of the Metal
We know:
- Mass of metal ((m_m)) = 7.00 g
- Initial temperature of metal ((T_{m,i})) = 93.0°C
- Mass of water ((m_w)) = 25.0 g
- Initial temperature of water ((T_{w,i})) = 30.0°C
- Final temperature ((T_f)) = 33.0°C
- Heat capacity of water = 1 cal/g°C
- Heat capacity of calorimeter ((C_{cal})) = 3.2 cal/°C
- Specific heat of metal ((c_m)) = ?
Step 1: Heat gained by water
[
q_{\text{water}} = m_w \cdot c \cdot (T_f – T_{w,i})
]
[
q_{\text{water}} = 25.0 \cdot 1 \cdot (33.0 – 30.0) = 25.0 \cdot 3.0 = 75.0 \, \text{cal}
]
Step 2: Heat gained by calorimeter
[
q_{\text{calorimeter}} = C_{cal} \cdot (T_f – T_{w,i})
]
[
q_{\text{calorimeter}} = 3.2 \cdot (33.0 – 30.0) = 3.2 \cdot 3.0 = 9.6 \, \text{cal}
]
Step 3: Heat lost by metal
[
q_{\text{metal}} = q_{\text{water}} + q_{\text{calorimeter}}
]
[
q_{\text{metal}} = 75.0 + 9.6 = 84.6 \, \text{cal}
]
Step 4: Specific heat of metal
[
q_{\text{metal}} = m_m \cdot c_m \cdot (T_{m,i} – T_f)
]
[
84.6 = 7.00 \cdot c_m \cdot (93.0 – 33.0)
]
[
84.6 = 7.00 \cdot c_m \cdot 60.0
]
[
c_m = \frac{84.6}{7.00 \cdot 60.0} = \frac{84.6}{420.0} = 0.20 \, \text{cal/g°C}
]
Explanation
In both problems, the principle of conservation of energy is applied. The heat lost by a warmer body equals the heat gained by cooler bodies (water and calorimeter). For problem 1, the calorimeter’s heat capacity is derived from the total heat balance equation. For problem 2, the specific heat of the metal is determined by isolating the heat balance and accounting for the calorimeter’s contribution.