1-0 kg of water is contained in a 1-25 kW kettle- Calculate the time taken for the temperature of water to rise from 25-C to its boiling point 100-C- Specific heat capacity of water - 4-20 J g-1 K-1 -

The correct option is C 252 s Heat required is Q= ms Δ T 1000 × 4.2× (100 25)=315000 J so the time required by heater of power 1250W to su ...

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Byju's Answer

Standard XII

Physics

Conduction Law

1.0 kg of wat...

Question

$1.0 k g$ of water is contained in a $1.25 k W$ kettle. Calculate the time taken for the temperature of water to rise from $25^{\circ} C$ to its boiling point $100^{\circ} C$ . Specific heat capacity of water = $4.20 J g^{- 1} K^{- 1}$ :

126 s

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252 s

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504 s

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none of the above

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Solution

The correct option is C. $252 s$ .
Heat required is $Q = m s Δ T$
$Q = 1000 \times 4.2 \times (100 - 25) = 315000 J$
so the time required by heater of power $1250 W$ to supply this amount of heat is
$t = \frac{Q}{P} = \frac{315000}{1250} = 252 s e c$ .
$∵ P = \frac{E n e r g y}{T i m e}$

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1.0 k g

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Q. 200 g of hot water at

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An electric kettle raises the temperature of the certain mass of water from 0 $^{\circ} C$ to 100 $^{\circ} C$ in 2.1 minutes. It takes 11.2 minutes to boil off the water. Calculate the specific heat of vaporization.

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1.0 kg of water is contained in a 1.25 kW kettle. Calculate the time taken for the temperature of water to rise from 25∘C to its boiling point 100∘C. Specific heat capacity of water = 4.20Jg−1K−1 :

The correct option is C. 252 s.Heat required is Q=msΔTQ=1000×4.2×(100−25)=315000 Jso the time required by heater of power 1250W to supply this amount of heat ist=QP=3150001250=252 sec.∵P=EnergyTime

$1.0 k g$ of water is contained in a $1.25 k W$ kettle. Calculate the time taken for the temperature of water to rise from $25^{\circ} C$ to its boiling point $100^{\circ} C$ . Specific heat capacity of water = $4.20 J g^{- 1} K^{- 1}$ :

The correct option is C. $252 s$ .
Heat required is $Q = m s Δ T$
$Q = 1000 \times 4.2 \times (100 - 25) = 315000 J$
so the time required by heater of power $1250 W$ to supply this amount of heat is
$t = \frac{Q}{P} = \frac{315000}{1250} = 252 s e c$ .
$∵ P = \frac{E n e r g y}{T i m e}$