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?


1 Answer 1


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

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

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

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

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


$Q = 150\times 334 = 50100kJ$

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

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

$Q = mc\delta T$


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


$\delta T = 1.76$

Therefore final temperature is 1.76 degrees Celsius.

  • $\begingroup$ 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? $\endgroup$
    – Cheryl
    Mar 22, 2013 at 4:43
  • $\begingroup$ @Cheryl, please see the edited answer to see how to solve the problem. $\endgroup$
    – Kenshin
    Mar 22, 2013 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.