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If $55 034.175 \rm{kJ }$ of heat are transferred to $150 \rm{kg}$ of ice at a temperature of $-12.15 ^\circ \rm{C}$, calculate the temperature of the resulting water.

Using $Q = mc(t_2-t_1)$ or $t_2 = \frac{Q}{mc}+t_1$,

$$t2 = \frac{55034.175}{150 \times 2.135} + (-12.15)$$

My answer is $178.6 ^\circ \rm{C}$

However, this does not seem possible to me because the boiling point of water is $100^\circ\rm{C}$ so therefore it would no longer be water. I feel my answer should be below $100^\circ\rm{C}$ if they are asking for the temperature of the resulting water.

Can someone please explain?

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1 Answer 1

First think how much energy causes a temperature change from -12.15 to 0 degrees Celcius.

$Q = mc\delta T = 150 \times 2.1 \times 12.15 = 3827.25 kJ$

We now have 55,034.175 - 3827 = 51207Kj remaining.

Now Ice is converted to water. The energy requried to convert Ice to Water is given by:

$Q = mL_f$, where $L_f$ is the latent heat of fusion for water, which is 334.

Therefore,

$Q = 150\times 334 = 50100kJ$

We therefore now have 51207-50100 = 1107kJ remaining.

So to determine how much this raises the temperature from 0, we use the formula,

$Q = mc\delta T$

Therefore,

$1107 = 150\times 4.2 \times \delta T$

Therefore

$\delta T = 1.76$

Therefore final temperature is 1.76 degrees Celsius.

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Thanks for your help but wouldn't you use a constant of 2.135 for a solid rather than 4.2 which is for a liquid? –  Cheryl Mar 22 '13 at 4:43
    
@Cheryl, please see the edited answer to see how to solve the problem. –  Mew Mar 22 '13 at 9:38
    
Thanks Chris...I appreciate your help. –  Cheryl Mar 23 '13 at 6:24
    
No problem, please accept my answer as best. –  Mew Mar 23 '13 at 6:54

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