a resistor has a voltage of 6.65 V and a current of 4.45 A. it is placed in a calorimeter containing 200g of water at 24 degree Celsius. the calorimeter is aluminum (specific heat= 0.220 cal/g-Celsius), and its mass is 60g. the heat capacity of the resistor itself is negligible. what is the temperature of the system 500 s later if all the electrical energy goes into heating the water and the calorimeter?
Please show me steps. The lab shows some equations i can use, but I do not fully understad.
P= VI = 29.6 watts
E= P*t = 14800 j
4.186 J = 1 c
So E= 14800/4.186 cal = 3535.6 cal
Q = c (specific heat) * m * T
T= Q/(m*C) = 3535.6/260* 0.220 = 61.8 C
T = Tf - ti
tf = 61.8 + 24 = 84.8 C
To solve this problem, we can use the formula for heat transfer:
Q = mcΔT
where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
First, let's calculate the heat transferred to the water and the calorimeter:
Q_water = mc_waterΔT_water
Q_calorimeter = mc_calorimeterΔT_calorimeter
Next, we need to find the initial temperature of the water and the final temperature of the system.
Given:
Voltage (V) = 6.65 V
Current (I) = 4.45 A
Time (t) = 500 s
Water mass (m_water) = 200 g
Initial water temperature (T_water_initial) = 24 °C
Calorimeter mass (m_calorimeter) = 60 g
Calorimeter specific heat (c_calorimeter) = 0.220 cal/g-°C
1. Calculate the heat transferred to the water:
Q_water = mc_waterΔT_water
= (200 g)(0.220 cal/g-°C)(ΔT_water)
2. Calculate the heat transferred to the calorimeter:
Q_calorimeter = mc_calorimeterΔT_calorimeter
= (60 g)(0.220 cal/g-°C)(ΔT_calorimeter)
3. Calculate the electrical energy used:
Electrical energy (E) = VIt
= (6.65 V)(4.45 A)(500 s)
4. Since the electrical energy used is equal to the heat transferred to water and calorimeter, we can equate the two:
E = Q_water + Q_calorimeter
5. Substitute the formulas for Q_water and Q_calorimeter, and solve for the change in temperature (ΔT_water + ΔT_calorimeter):
(6.65 V)(4.45 A)(500 s) = (200 g)(0.220 cal/g-°C)(ΔT_water) + (60 g)(0.220 cal/g-°C)(ΔT_calorimeter)
6. Rearrange the equation to solve for ΔT_water + ΔT_calorimeter:
(ΔT_water) + (ΔT_calorimeter) = (6.65 V)(4.45 A)(500 s) / [(200 g)(0.220 cal/g-°C) + (60 g)(0.220 cal/g-°C)]
7. Calculate ΔT_water + ΔT_calorimeter using the formula above.
8. Finally, calculate the final temperature of the system by adding ΔT_water + ΔT_calorimeter to the initial temperature of the water:
T_system_final = T_water_initial + (ΔT_water + ΔT_calorimeter)
Substitute the values and calculate to find the final temperature of the system.
To find the final temperature of the system after 500 seconds, we'll apply the principle of conservation of energy.
The electrical energy supplied to the resistor can be calculated using the formula:
Electrical Energy = Voltage x Current x Time
Given:
Voltage (V) = 6.65 V
Current (I) = 4.45 A
Time (t) = 500 s
Electrical Energy = 6.65 V x 4.45 A x 500 s = 14,787.25 Joules (J)
Since the electrical energy goes into heating the water and the calorimeter, we can use the formula for heat energy:
Heat Energy = Mass x Specific Heat x Temperature Change
We need to calculate the temperature change (∆T) in the system. The heat energy gained by the water and the calorimeter equals the electrical energy supplied:
Mass of Water (m) = 200 g
Specific Heat of Aluminum (C_aluminum) = 0.220 cal/g-°C
Mass of Calorimeter (m_calorimeter) = 60 g
Heat Energy = (m + m_calorimeter) x Specific Heat x ∆T
Substituting the given values:
14,787.25 J = (200 g + 60 g) x 0.220 cal/g-°C x ∆T
To convert joules to calories, we'll divide the energy by the conversion factor 1 J = 0.239 cal:
14,787.25 J ÷ 0.239 cal/J = 61,796.15 cal
Now, solving for ∆T:
61,796.15 cal = (200 g + 60 g) x 0.220 cal/g-°C x ∆T
61,796.15 cal = 260 g x 0.220 cal/g-°C x ∆T
61,796.15 cal = 57.2 cal/°C x ∆T
∆T = 61,796.15 cal ÷ 57.2 cal/°C ≈ 1,080.25 °C
Since we're given the initial temperature of the water as 24 °C, the final temperature of the system will be:
Final Temperature = Initial Temperature + ∆T
Final Temperature = 24 °C + 1,080.25 °C ≈ 1,104.25 °C
Therefore, the temperature of the system 500 s later, if all the electrical energy goes into heating the water and the calorimeter, is approximately 1,104.25 °C.