Question

How much heat transfer is required to raise the temperature of a $0.45kg$ aluminium pot containing $1.5kg$ of ice from $-{10}^{o}C$ to boiling point and then boil away the water? How long does it take if the rate of heat transfer is $300W$?

Answer 1

Initial temperature of pot + ice=$-10°C$

Heat required to convert ice from,

-10°C to 0°C= mc$\Delta$T=$Q_1$

Where,

m=Mass

C=Thermal capacity of ice

$\Delta$T=Temperature difference

Thermal capacity of ice= 0.5cal/g°C

$Q_1=mc\Delta T=1500\times \frac{1}{2}\times 10=7,500$calories

Heat required to change the state of ice $=Q_2$=mL

Where, m=Mass, L=Latent heat

$Q_2=1500\times 80=1,20,000$calories

Heat required to change the temperature of water(melted ice) $=Q_3=mc\Delta T$

$Q_3=1500\times 1\times 100=1,50,000$ calories

Heat required to change the state of water= mL=$Q_4$

$Q_4=1500\times 540$ = $8,10,000$ calories

Heat required to change the temperature of aluminum pot $=Q_5=mc\Delta T$

Specific heat capacity of aluminum =0.21cal/g°C

$Q_5=mc\Delta T$

$450\times 0.21\times 110$ = $10,395$ calories

Total heat required $=Q_1+Q_2+Q_3+Q_4+Q_5$ = $10,97,895$ calories = $46,11,159$ Joules

We are given $P=300W=300Joulespersecond$

300 Joules are produced in = 1 s

1 Joule is produced in $=\frac{1}{300}$ s

46,11,159 Joules is produced in $=\frac{1}{300}\times 46,11,159$

=$15,370.53$ seconds