Two litre of water at initial temperature of 300k is heated by a heater of power 1kW. if the lid of kettle is opened then heat is losted at the constant rate of 160J/s.Find the time required to raise the temperature of water to 350k with the lid open(Specific heat of water 4.2 kJ/kg) ...................................................................................................
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To solve the problem of heating 2 liters of water from 300 K to 350 K with a heater and heat loss, we need to consider both the energy supplied by the heater and the energy lost due to heat escaping. Let's break this down step by step.

Understanding the Variables

First, we need to identify the key variables in the problem:

• Volume of water: 2 liters (which is equivalent to 2 kg, since the density of water is approximately 1 kg/L).
• Initial temperature (T1): 300 K.
• Final temperature (T2): 350 K.
• Power of the heater: 1 kW (which is 1000 J/s).
• Heat loss rate: 160 J/s.
• Specific heat capacity of water (c): 4.2 kJ/kg·K (which is 4200 J/kg·K).

Calculating the Energy Required

Next, we need to calculate the total energy required to raise the temperature of the water:

The formula for the energy required (Q) to change the temperature of a substance is:

Q = mcΔT

Where:

• m: mass of the water (2 kg)
• c: specific heat capacity (4200 J/kg·K)
• ΔT: change in temperature (T2 - T1 = 350 K - 300 K = 50 K)

Plugging in the values:

Q = 2 kg × 4200 J/kg·K × 50 K

Q = 420000 J

Calculating the Effective Power

Now, we need to consider the effective power of the heater after accounting for the heat loss. The net power available for heating the water is:

Net Power = Power of Heater - Heat Loss Rate

Net Power = 1000 J/s - 160 J/s = 840 J/s

Finding the Time Required

To find the time (t) required to supply the necessary energy (Q) with the net power, we can use the formula:

t = Q / Net Power

Substituting the values we calculated:

t = 420000 J / 840 J/s

t = 500 seconds

Final Thoughts

Therefore, the time required to raise the temperature of the water to 350 K with the lid open is 500 seconds. This calculation illustrates the importance of considering both the energy input and the energy losses in thermal systems.