A 500g aluminum pot containing 250mL of water in the pot increases from 13.2 °C to 89.1 °C. Calculated the heat absorbed by the pot and water.
I don't know the official approach to this question, somebody help!
Heatabsorbed= masspot*cpot*deltaTemp+masswater*cwater*deltatemp
That is the official way.
To calculate the heat absorbed by the pot and water, we can use the formula:
Q = m * c * ΔT
Where:
- Q is the heat absorbed or released (in joules)
- m is the mass of the substance (in grams or kilograms)
- c is the specific heat capacity of the substance (in J/g°C or J/kg°C)
- ΔT is the change in temperature (in °C)
Let's calculate the heat absorbed by the pot and water separately:
1. Heat absorbed by the pot:
The mass of the pot is given as 500g, and assuming the pot is made of aluminum, the specific heat capacity of aluminum is 0.897 J/g°C.
Using the formula Q = m * c * ΔT, we can calculate the heat absorbed by the pot:
Q_pot = 500g * 0.897 J/g°C * (89.1 °C - 13.2 °C)
= 500g * 0.897 J/g°C * 75.9 °C
= 33,967.5 J (or 33.97 kJ)
Therefore, the heat absorbed by the pot is 33,967.5 Joules (or 33.97 kilojoules).
2. Heat absorbed by the water:
The volume of water is given as 250mL, which can be converted to grams assuming the density of water is 1g/mL.
Using the formula Q = m * c * ΔT, we need to find the mass of the water first:
mass_water = volume_water * density_water
= 250mL * 1g/mL
= 250g
The specific heat capacity of water is approximately 4.18 J/g°C.
Using the formula Q = m * c * ΔT, we can calculate the heat absorbed by the water:
Q_water = 250g * 4.18 J/g°C * (89.1 °C - 13.2 °C)
= 250g * 4.18 J/g°C * 75.9 °C
= 75,947.5 J (or 75.95 kJ)
Therefore, the heat absorbed by the water is 75,947.5 Joules (or 75.95 kilojoules).
So, the heat absorbed by the pot and water combined is the sum of these two values:
Q_total = Q_pot + Q_water
= 33,967.5 J + 75,947.5 J
= 109,915 J (or 109.92 kJ)
Therefore, the total heat absorbed by the pot and water is 109,915 Joules (or 109.92 kilojoules).