# Sensible heat question (solving for temperature)

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.

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.

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.

• 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. – Kenshin Mar 22 '13 at 9:38