a heater rated 1000 W is used to heat 1.6 kg of water at 40°c to its boiling point. Calculate the time in which water starts to boil. Specific heat capacity of water is 4200J/kg/°c.

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Solution

Dear Student,

$Energy supplied = P \times t . . . . . . . . . . . . . . . . . . . . . . . . . . (1) Energy absorbed = {mCθ}_{R} Energy absorbed = 1.6 \times 4200 \times (100 - 40) Energy absorbed = 403200 J As, energy supplied = energy absorbed P \times t = 403200 t = \frac{403200}{1000} = 403.2 s$

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25^{\circ} C

to its boiling point

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. Specific heat capacity of water =

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100^{o} C

. Specific heat capacity of water

= 4.2 J g^{- 1} K^{- 1}

M kg

of water at

20^{\circ} C

is brought to its boiling point. If the heat energy gained by

M kg

of water is

3360 kJ

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M

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Heat is supplied at the same rate to equal masses of ice and water at the melting point (MP) of ice. Calculate the ratio of the time required to melt the ice and to bring the water to its Boiling Point. [Given that specific heat capacity of water = 4.2 J/g^oC, Specific latent heat of ice = 336 J/g]

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a heater rated 1000 W is used to heat 1.6 kg of water at 40°c to its boiling point. Calculate the time in which water starts to boil. Specific heat capacity of water is 4200J/kg/°c.

Dear Student, Energy supplied=P×t .......................... 1Energy absorbed=mCθREnergy absorbed=1.6×4200×100-40Energy absorbed=403200 JAs, energy supplied=energy absorbedP×t=403200t=4032001000=403.2 s