A piece of metal of mass 400g of specific heat capacity 500jkgk is place in an oven to attain a study temperature and transfer in to a calorimeter with 1000g of water at 30c the thermal capacity of calorimeter and stirrer is 105jk final temperature of the mixture 35c fine the temperature of the oven
calorimeter and water goes up from 30 to 35
heat in
105 J/degK (35-30) + 1.00 kg * specific heat of water * (35-30)
heat out
0.400 * 500 J/ kg deg K ( T - 35)
heat in = heat out, solve for T
To find the temperature of the oven, we can use the principle of conservation of energy. The heat gained by the metal in the oven is equal to the heat lost by the metal and water in the calorimeter.
Let's calculate the heat gained by the metal in the oven:
1. Mass of the metal (m) = 400g = 0.4 kg
2. Specific heat capacity of the metal (c) = 500 J/kgK
3. Initial temperature of the metal (t1) = ?
4. Final temperature of the metal (t2) = 35°C (converted to Kelvin: T2 = 35 + 273 = 308 K)
The heat gained by the metal in the oven can be calculated using the formula:
Q1 = mcΔT1
where Q1 is the heat gained by the metal, m is the mass of the metal, c is the specific heat capacity, and ΔT1 is the change in temperature.
Since the temperature in the oven is constant, the change in temperature (ΔT1) is:
ΔT1 = t2 - t1
Substituting the given values:
ΔT1 = 308 K - t1
Thus, the heat gained by the metal in the oven is:
Q1 = mcΔT1
= (0.4 kg) * (500 J/kgK) * (308 K - t1)
Now, let's calculate the heat lost by the metal and water in the calorimeter:
1. Mass of water (m2) = 1000g = 1 kg
2. Specific heat capacity of water (c2) = 4186 J/kgK
3. Initial temperature of water (t2) = 30°C (converted to Kelvin: T2 = 30 + 273 = 303 K)
4. Final temperature of the mixture (t3) = 35°C (converted to Kelvin: T3 = 35 + 273 = 308 K)
5. Thermal capacity of the calorimeter and stirrer (C) = 105 J/K
The heat lost by the metal and water in the calorimeter can be calculated using the formula:
Q2 = m2c2ΔT2
where Q2 is the heat lost by the metal and water, m2 is the mass of water, c2 is the specific heat capacity of water, and ΔT2 is the change in temperature.
Since the temperature in the calorimeter is constant, the change in temperature (ΔT2) is:
ΔT2 = t3 - t2
= 308 K - 303 K
Thus, the heat lost by the metal and water in the calorimeter is:
Q2 = m2c2ΔT2
= (1 kg) * (4186 J/kgK) * (308 K - 303 K)
According to the principle of conservation of energy:
Q1 = Q2
So,
(0.4 kg) * (500 J/kgK) * (308 K - t1) = (1 kg) * (4186 J/kgK) * (308 K - 303 K)
Simplifying the equation:
(0.4 * 500 * 308) - (0.4 * 500 * t1) = (1 * 4186 * 5)
Solving for t1:
(0.4 * 500 * t1) = (0.4 * 500 * 308) - (1 * 4186 * 5)
t1 = [(0.4 * 500 * 308) - (1 * 4186 * 5)] / (0.4 * 500)
Using a calculator, t1 approximately equals to 468.6°C.
Therefore, the temperature of the oven is approximately 468.6°C.
To find the temperature of the oven, we can use the principle of energy conservation.
First, let's break down the problem into two steps:
Step 1: The metal piece achieving a steady temperature in the oven.
Step 2: The transfer of heat from the metal to the water in the calorimeter.
Step 1: Finding the steady temperature of the metal in the oven.
To find the steady temperature, we can use the formula:
Q = mcΔT
where Q is the heat gained or lost, m is the mass of the metal, c is the specific heat capacity of the metal, and ΔT is the change in temperature.
In this case, the heat gained by the metal is equal to the heat lost by the oven. Therefore:
Q_metal = Q_oven
Using the formula, we find:
mcΔT_metal = mcΔT_oven
Given that the mass of the metal (m) is 400 g, the specific heat capacity of the metal (c) is 500 J/kg·K, and the change in temperature of the oven (ΔT_oven) is unknown, we can rearrange the formula to solve for ΔT_oven:
ΔT_oven = (mcΔT_metal) / mc
Step 2: Transfer of heat from the metal to the water in the calorimeter.
To calculate the heat transferred from the metal to the water, we can use the formula:
Q = mcΔT
where Q is the heat gained or lost, m is the mass of the substance (in this case, water), c is the specific heat capacity of the substance, and ΔT is the change in temperature.
The heat gained by the water is equal to the heat lost by the metal. Therefore:
Q_water = Q_metal
Using the formula, we find:
mcΔT_water = mcΔT_metal
Given that the mass of the water (m) is 1000 g, the specific heat capacity of water (c) is approximately 4186 J/kg·K, and the change in temperature of the water (ΔT_water) is (35°C - 30°C), we can rearrange the formula to solve for ΔT_metal:
ΔT_metal = (mcΔT_water) / mc
Finally, using the values obtained from Step 1 and Step 2, we substitute them into the equation we derived in Step 1 to find the temperature of the oven:
ΔT_oven = (mcΔT_metal) / mc
Substituting the calculated values will give us the temperature of the oven.