2 kg of water is heated with a heating coil which draws 3.5 A from a 200 V mains for 2 minutes. what is the increase in temperature of the water? [specific ...
Question 1 Report
2 kg of water is heated with a heating coil which draws 3.5 A from a 200 V mains for 2 minutes. what is the increase in temperature of the water?
[specific heat capacity of water = 4200 jkg -1k-1]
[specific heat capacity of water = 4200 jkg -1k-1]
Answer Details
The increase in temperature of the water can be calculated using the formula:
Q = mcΔT
Where Q is the heat energy transferred to the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
First, let's calculate the heat energy transferred to the water:
Power = VI = 200V × 3.5A = 700W
Energy = Power × time = 700W × 2 minutes = 84000J
Now, let's calculate the mass of the water:
1 kg of water has a volume of 1 liter, and the density of water is 1000 kg/m³. Therefore, 2 kg of water has a volume of 2 liters or 0.002 m³.
Next, we can use the formula to calculate the change in temperature:
ΔT = Q / (mc)
ΔT = 84000J / (2kg × 4200J/kg.K)
ΔT = 10°C
Therefore, the increase in temperature of the water is 10°C. Option (d) is the correct answer.
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