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2 litres water at 27^(@)C is heated by a...

`2` litres water at `27^(@)C` is heated by a `1kW` heater in an open container. On an average heat is lost to surroundings at the rate `160 J//s`. The time required for the temperature to reach `77^(@)C` is

A

`8 min 20 sec`

B

`10 min`

C

`7 min`

D

`14 min`

Text Solution

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The correct Answer is:
To solve the problem step by step, we will follow the process of calculating the effective power of the heater, the heat required to raise the temperature of the water, and finally, the time taken to reach the desired temperature. ### Step 1: Calculate the effective power of the heater The heater has a power of 1 kW, which is equivalent to 1000 Joules per second. However, it loses heat to the surroundings at a rate of 160 Joules per second. Therefore, we need to calculate the effective power that contributes to heating the water. \[ \text{Effective Power} = \text{Power of Heater} - \text{Heat Lost} \] \[ \text{Effective Power} = 1000 \, \text{J/s} - 160 \, \text{J/s} = 840 \, \text{J/s} \] ### Step 2: Calculate the heat required to raise the temperature of the water We need to calculate the heat required to raise the temperature of 2 liters of water from 27°C to 77°C. First, convert the volume of water to mass: \[ \text{Mass of water} = 2 \, \text{liters} = 2000 \, \text{grams} \] Next, calculate the change in temperature: \[ \Delta T = 77°C - 27°C = 50°C \] Now, using the specific heat capacity of water, which is approximately \(4.2 \, \text{J/g°C}\), we can calculate the total heat required (Q): \[ Q = \text{mass} \times \text{specific heat} \times \Delta T \] \[ Q = 2000 \, \text{g} \times 4.2 \, \text{J/g°C} \times 50°C \] \[ Q = 2000 \times 4.2 \times 50 = 420000 \, \text{J} \] ### Step 3: Calculate the time required to reach the desired temperature Now, we can find the time (T) required to supply the necessary heat using the effective power calculated earlier. The formula relating power, time, and energy is: \[ \text{Power} = \frac{Q}{T} \implies T = \frac{Q}{\text{Effective Power}} \] Substituting the values we have: \[ T = \frac{420000 \, \text{J}}{840 \, \text{J/s}} = 500 \, \text{s} \] ### Step 4: Convert time from seconds to minutes To convert seconds into minutes: \[ T = \frac{500 \, \text{s}}{60} \approx 8.33 \, \text{minutes} \] This is approximately 8 minutes and 20 seconds. ### Final Answer The time required for the temperature to reach 77°C is approximately **8 minutes and 20 seconds**. ---

To solve the problem step by step, we will follow the process of calculating the effective power of the heater, the heat required to raise the temperature of the water, and finally, the time taken to reach the desired temperature. ### Step 1: Calculate the effective power of the heater The heater has a power of 1 kW, which is equivalent to 1000 Joules per second. However, it loses heat to the surroundings at a rate of 160 Joules per second. Therefore, we need to calculate the effective power that contributes to heating the water. \[ \text{Effective Power} = \text{Power of Heater} - \text{Heat Lost} \] ...
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Water of volume 2 litre in a container is heated with a coil of 1 kW at 27^(circ) C . The lid of the container is open and cnergy dissipates at rate of 160 J / s . In how much time (in s ) temperature will rise from 27^(circ) C to 77^(circ) C ? (Given: Specific heat of water is 4.2 kJ / kg)

Knowledge Check

  • A kettle with 2 litre water at 27^@C is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate 160 J//s , when its lid is open. In how much time will water heated to 77^@C with the lid open ? (specific heat of water = 4.2 kJ//^@C-kg)

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    B
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    C
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    D
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    B
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  • Two litres of water kept in a container at 27^(@)C is heated with a coil of 1 kW. The lid of the container is open and energy dissipates at the rate of 160 J/s. If the specific heat of water is 4.2 kJ/kg, then time taken by coil to raise the temperature of water from 27^(@)C to 77^(@)C is

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