# Finding the final temperature

If $125$ Calories of heat is applied to $60.0\text{g}$ piece of copper at $21.0^\circ\text{C}$, what will the final temperature be? The specific head of copper is $0.0920 \text{ cal/gC}$.

All I have to answer this question is a formula:

$SH = \text{heat/(mass}\times \Delta T)$

• Please see meta.chemistry.stackexchange.com/questions/141/… for how to ask homework questions. Please outline what you have attempted towards solving the problem. – jonsca Feb 9 '14 at 19:13
• Any hint would be great. I'm really not sure how to go about the answer. I've been working on it for about 3 hours already, as silly as it sounds. – ninja08 Feb 9 '14 at 19:16
• Rather than just us being draconian, it is helpful for those answering to see what your thought process is so they can steer you in the right direction. What else do you know about the problem? Do you have a formula in mind, if so, do the dimensions work out? – jonsca Feb 9 '14 at 19:19
• Split $\Delta T$ up into $T_f - T_i$, and you have all of the rest of the parameters. – jonsca Feb 9 '14 at 19:24
• And now, post your answer below so the next student finds your method instead of having to ask another question ;) – tschoppi Feb 10 '14 at 18:49

The energy amount used for heating up a material can be calculated according to the following equation: $$E = c_\text{s}\, m\, \Delta T$$ where $E$ is the energy, $c_\text{s}$ the specific heat capacity, $m$ the mass of the object and $\Delta T$ the temperature difference that is to be achieved.
Solving the above equation for $\Delta T$ we get: $$\Delta T = \frac{E}{c_\text{s}\, m}$$
Plugging in the values provided by you, I get $\Delta T = 22.6\;\mathrm{K}$ (or °C, since we're talking about the difference and the scale is the same, it doesn't matter).
Adding this increase in temperature to the initial temperature gives the final answer: $$\vartheta_\text{f} = \vartheta_\text{i} + \Delta \vartheta = 21.0\; \mathrm{°C} + 22.6\; \mathrm{°C} = 43.6\; \mathrm{°C}$$
• Symbol for temperature: I used $\vartheta$ for temperatures in °C and $T$ for temperatures in K. I strongly suggest you follow this example as to never confuse the two.