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(a) how much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from 30.0ºc to the boiling point and then boil away 0.750 kg of water? (b) how long does this take if the rate of heat transfer is 500 w 1 watt = 1 joule/second (1 w = 1 j/s)?

Respuesta :

Heat required to raise the temperature is given by formula

[tex] Q = mc\Delta T[/tex]

here

[tex]c_{aluminium} = 900[/tex]

[tex]c_{water} = 4186[/tex]

[tex]L_f = 2.25 * 10^6[/tex]

now here first temperature of water and aluminium container raised to 100 degree C then water is boiled away.

heat is calculated by formula

[tex]Q = m_w C_w \Delta T + m_A C_A \Delta T + m_w*L_f[/tex]

[tex]Q = 2.50* 4186* (100 - 30) + 0.750* 900* (100 - 30) + 0.750 * 2.25 * 10^6[/tex]

[tex]Q = 732550 + 47250 + 1.68*10^6[/tex]

[tex]Q = 2.46* 10^6 J[/tex]

PART b)

If power supplied is 500 W

then time taken to give such amount of heat is

[tex] t = \frac{Q}{P}[/tex]

[tex] t = \frac{2.468* 10^6}{500}[/tex]

[tex]t = 1.37 Hours[/tex]

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