calculate the final temperature of a 400g of water at 80 degree celsius if it is poured into a 100g pot at 30 degree celsius. note: c of pot is 840 J/kg degree celsius.
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To calculate the final temperature, we can use the principle of conservation of energy.
First, let's calculate the heat gained by the pot. We can use the formula:
Q pot = m pot * c pot * ΔT
Where:
m pot = mass of the pot = 100g = 0.1 kg
c pot = specific heat capacity of the pot = 840 J/kg°C
ΔT = change in temperature = final temperature - initial temperature = Tf - 30°C
Next, let's calculate the heat lost by the water. We use the same formula:
Q water = m water * c water * ΔT
Where:
m water = mass of the water = 400g = 0.4 kg
c water = specific heat capacity of water = 4186 J/kg°C
ΔT = change in temperature = final temperature - initial temperature = 80°C - Tw
Since heat gained by the pot equals heat lost by the water, we can set up the equation:
Q pot = Q water
m pot * c pot * ΔT = m water * c water * ΔT
Substituting the given values:
0.1 kg * 840 J/kg°C * (Tf - 30°C) = 0.4 kg * 4186 J/kg°C * (80°C - Tf)
Simplifying the equation:
840 ( Tf - 30) = 0.4 * 4186 (80 - Tf)
840 Tf - 25,200 = 3344 - 0.4 Tf
840 Tf + 0.4 Tf = 3344 + 25,200
840.4 Tf = 28,544
Tf = 28,544 / 840.4
Tf ≈ 34°C (rounded to the nearest whole number)
Therefore, the final temperature of the system will be approximately 34 degrees Celsius.
To calculate the final temperature of the water and pot system, we can use the principle of conservation of energy. The total heat gained by the water and pot system must be equal to the total heat lost by the water and pot system.
Let's break down the calculation into two steps:
Step 1: Calculate the heat gained by the water.
The formula to calculate the heat gained by an object is given by:
Q = mcΔT
Where:
Q is the heat gained by the object (in joules)
m is the mass of the object (in kilograms)
c is the specific heat capacity of the object (in J/kg °C)
ΔT is the change in temperature of the object (in °C)
For the water, the mass (m) is 400g, which is 0.4kg, and the initial temperature (ΔT) is 80°C. Since water has a specific heat capacity of 4186 J/kg °C, we can calculate the heat gained by the water as follows:
Q_water = m_water * c_water * ΔT_water
= 0.4kg * 4186 J/kg °C * (80°C - T_final)
Step 2: Calculate the heat lost by the pot.
Similar to step 1, we can calculate the heat lost by the pot using the formula:
Q = mcΔT
For the pot, the mass (m) is 100g, which is 0.1kg, and the initial temperature (ΔT) is 30°C. The specific heat capacity of the pot (c_pot) is given as 840 J/kg °C. Therefore, the heat lost by the pot can be calculated as follows:
Q_pot = m_pot * c_pot * ΔT_pot
= 0.1kg * 840 J/kg °C * (T_final - 30°C)
Since the total heat gained by the water and pot system must be equal to the total heat lost, we have:
Q_water = Q_pot
0.4kg * 4186 J/kg °C * (80°C - T_final) = 0.1kg * 840 J/kg °C * (T_final - 30°C)
Now we can solve for T_final:
0.4 * 4186 * (80 - T_final) = 0.1 * 840 * (T_final - 30)
1674 * (80 - T_final) = 84 * (T_final - 30)
133920 - 1674T_final = 84T_final - 2520
1258T_final = 136440
T_final = 136440 / 1258
T_final ≈ 108.61°C
Therefore, the final temperature of the water and pot system is approximately 108.61°C.