SATHEE: Heat and Thermodynamics 1 Question 12

Heat and Thermodynamics 1 Question 12

12. Water of volume $2 L$ in a container is heated with a coil of $1 k W$ at $27^{\circ} C$ . The lid of the container is open and energy dissipates at rate of $160 J / s$ . In how much time temperature will rise from $27^{\circ} C$ to $77^{\circ} C$ ?

(2005, 2M)

[Specific heat of water is $4.2 k J / k g$ ]

(a) $8 m i n 20 s$

(b) $6 min 2 s$

(d) $14 m i n$

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Answer:

Correct Answer: 12. (a)

Solution:

Energy gained by water (in $1 s$ )

$=$ energy supplied - energy lost $= 1000 J - 160 J = 840 J$

Total heat required to raise the temperature of water from $27^{\circ} C$ to $77^{\circ} C$ is $m s Δ θ$ .

Hence, the required time

$\begin{aligned} t & = \frac{m s Δ θ}{rate by which energy is gained by water} \\ = \frac{(2) (4.2 \times 10^{3}) (50)}{840} \\ = 500 s = 8 m i n 20 s \end{aligned}$

Heat and Thermodynamics 1 Question 12

12. Water of volume $2 L$ in a container is heated with a coil of $1 k W$ at $27^{\circ} C$ . The lid of the container is open and energy dissipates at rate of $160 J / s$ . In how much time temperature will rise from $27^{\circ} C$ to $77^{\circ} C$ ?

Answer:

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Heat and Thermodynamics 1 Question 12

12. Water of volume 2L in a container is heated with a coil of 1kW at 27∘C. The lid of the container is open and energy dissipates at rate of 160J/s. In how much time temperature will rise from 27∘C to 77∘C ?

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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12. Water of volume $2 L$ in a container is heated with a coil of $1 k W$ at $27^{\circ} C$ . The lid of the container is open and energy dissipates at rate of $160 J / s$ . In how much time temperature will rise from $27^{\circ} C$ to $77^{\circ} C$ ?