A lead calorimeter with mass of 150g contains 200g of water. The calorimeter and water are in thermal equilibrium at a specific temperature. Two metallic blocks are placed into the water. One is a 50.0g piece of aluminum heated at 90 degrees celcius. The other has a mass of 65g and is originally at a temperature of 97 degrees celcius. The entire system reaches a final temperature of 25 degrees celcius. (a) Find the final temp of the calorimeter and water. (b) Determine the specific heat of the sample.
To solve this problem, we can use the principle of conservation of energy, which states that the heat gained by the water, calorimeter, and metallic blocks is equal to the heat lost by the aluminum and the metallic block.
Let's denote the final temperature of the water and calorimeter as T_f.
(a) To find the final temperature, we can use the equation:
(m_c * C_c + m_w * C_w) * (T_f - T_i) = m_A * C_A * (T_i - T_f) + m_B * C_B * (T_i - T_f)
where:
m_c = mass of the calorimeter (150g)
C_c = specific heat capacity of the calorimeter
m_w = mass of the water (200g)
C_w = specific heat capacity of water
m_A = mass of aluminum (50.0g)
C_A = specific heat capacity of aluminum
m_B = mass of the metallic block (65g)
C_B = specific heat capacity of the metallic block
T_i = initial temperature (25 degrees Celsius)
We know that the specific heat capacity of water is 4.18 J/g°C, and the specific heat capacity of aluminum is 0.90 J/g°C. However, we need the specific heat capacity of the metallic block (C_B) and calorimeter (C_c) to solve the equation.
To find the specific heat capacity of the metallic block (C_B), let's assume that it is the same as aluminum. We can then solve for the final temperature, T_f:
(150g * C_c + 200g * 4.18 J/g°C) * (T_f - 25°C) = 50.0g * 0.90 J/g°C * (90°C - T_f) + 65g * 0.90 J/g°C * (97°C - T_f)
Simplifying the equation, we have:
(150g * C_c + 836g * (T_f - 25°C)) = (45g * (90°C - T_f) + 58.5g * (97°C - T_f))
Expanding the equation further:
150g * C_c + 836g * T_f - 20900g = 4050g - 45g * T_f + 5584.5g - 58.5g * T_f
Combining like terms:
(150g * C_c + 836g * T_f) = -5584.5g + 45g * T_f - 58.5g * T_f + 20900g - 4050g
Simplifying:
(150g * C_c + 778.5g * T_f) = -144g + 13.5g * T_f + 16850g
(150g * C_c + 778.5g * T_f) = 16654g + 13.5g * T_f
Rearranging the equation:
778.5g * T_f - 13.5g * T_f = 16654g - 150g * C_c
765g * T_f = 16404g - 150g * C_c
T_f = (16404g - 150g * C_c) / 765g [Equation 1]
Now, to find the specific heat capacity of the calorimeter (C_c), we can use the fact that the calorimeter and water are in thermal equilibrium, which means that the heat lost by the aluminum and metallic block is equal to the heat gained by the calorimeter and water:
m_A * C_A * (T_i - T_f) + m_B * C_B * (T_i - T_f) = m_c * C_c * (T_f - T_i) + m_w * C_w * (T_f - T_i)
Substituting the known values:
(50.0g * 0.90 J/g°C * (25°C - T_f)) + (65g * 0.90 J/g°C * (25°C - T_f)) = 150g * C_c * (T_f - 25°C) + 200g * 4.18 J/g°C * (T_f - 25°C)
Simplifying:
45g * 0.90 J/g°C * (T_f - 25°C) + 58.5g * 0.90 J/g°C * (T_f - 25°C) = 150g * C_c * (T_f - 25°C) + 836g * (T_f - 25°C)
Combining like terms:
90.9g * (T_f - 25°C) = 150g * C_c * (T_f - 25°C) + 836g * (T_f - 25°C)
Expanding:
90.9g * T_f - 2272.5g = 150g * C_c * T_f - 3750g * C_c + 836g * T_f - 20900g
Combining like terms:
90.9g * T_f - 836g * T_f = 2272.5g - 3750g * C_c + 20900g
-745.1g * T_f = 18690.5g - 3750g * C_c
T_f = (18690.5g - 3750g * C_c) / -745.1g [Equation 2]
Now, we have two equations (Equation 1 and Equation 2) with two unknowns (C_c and T_f). We can solve this system of equations simultaneously to find the values.
(b) Once we have determined the final temperature of the water and calorimeter, we can find the specific heat of the sample using the equation:
Q = m_x * C_x * (T_x - T_f)
We need to use either the aluminum or the metallic block for this calculation. Let's use the aluminum:
Q = 50.0g * C_A * (90°C - T_f)
where Q is the heat gained by the water, calorimeter, and metallic block (which is equal to the heat lost by the aluminum).
Now, substitute the final temperature we found in part (a) into the equation and solve for C_A:
Q = 50.0g * 0.90 J/g°C * (90°C - T_f)
Finally, calculate the value for Q (heat gained) using the specific heat capacities and final temperature:
Q = (m_c * C_c + m_w * C_w) * (T_f - T_i)