Question

A metallic calorimeter of mass 100g contains 200 g of ice and the initial temperature of the calorimeter and the ice is $-{10}^{\circ }C$. Heat is supplied to the system containing calorimeter and ice at a constant rate of $50cal{s}^{-1}$. Find the time required to raise the temperature of the system to ${50}^{\circ }C$. [Neglect the loss of heat to the surroundings].

Take specific heat capacity of calorimeter and ice as $0.2cal{g}^{-1}{}^{\circ }{C}^{-1}$ and $\frac{1}{2}cal{g}^{-1}{}^{\circ }{C}^{-1}$.Take latent heat of fusion of ice as $80cal{g}^{-1}$ and specific heat capacity of water as $1cal{g}^{-1}{}^{\circ }{C}^{-1}$

Answer 1

Given data:

$\to$ Mass of calorimeter = 100gm

$\to$ Mass of ice =200gm

$\to$ initial temperature of calorimeter + ice $={-10}^{o}C$

$\to$ Rate of heat supplied to calorimeter$=50{cals}^{-1}$

$\to$ Final temperature of system(calorimeter+ice)$={50}^{o}C$

$\to$ Specific heat of calorimeter $=0.2cal{g}^{-1}{{}^{o}C}^{-1}$

$\to$ specific heat of ice $=\frac{1}{2}cal{g}^{-1}{{}^{o}C}^{-1}$

$\to$ Latent heat of fusion $=80cal/gm$

$\to$ Specific heat capacity of water $=1cal{g}^{-1}{{}^{o}C}^{-1}$

Now in the question:

$\to$ temperature of the system is changed from $-{10}^{o}C$ to ${50}^{o}C$

$\to$ It means heat is supplied to the system for

(1) heat required to change the temperature of calorimeter from $-{10}^{o}C$ to ${50}^{o}C$

Now

$Q_1=mc△T$

$=100\times 0.2\times 60=1200calories$

(2) heat required to change the temperature of ice from $-{10}^{o}Cto{0}^{o}C$

heat required $=mc\Delta T$

$Q_2=200\times \frac{1}{2}\times 10=1000calories$

(3) Heat required to change the state of ice i.e from solid to liquid (ice converted to water)

$Q_3=mL$

where L =latent of heat

$Q_3=200\times 80=16000calories$

(4) Heat required to change the temperature of water from $0C^{0}to{50}^{o}C$

$Q_4=mc\Delta T$ = $200\times 1\times 50=10000calories$

Total heat needed to convert the temperature of substance$=Q_1+Q_2+Q_3+Q_4$

$Q=1200+1000+16000+10000=28200calories$

But we are given rate of heat supplied =$50cal/s$

$50$ calories are supplied in = $1s$

$1$ calorie is supplied in $=\frac{1}{50}s$

$28200$ calories are supplied in $=(\frac{1}{50}\times 28200)s=564seconds$

Hence, the answer is $564$ seconds