24.5WTF2 2.5 x 10^{1}
WTF3 25
Step 1: Solve using the equation: $$ \Delta T = \dfrac{q}{m \cdot s} $$
Step 2: Using a bar problem: $$ \dfrac{26,000~J}{} \times \dfrac{1}{630 ~g} \times \dfrac{g^{\circ}C}{1.687 ~J} = {25 ~^{\circ} C \atop ~or~ 2.5 x 10^{1} ~^{\circ} C }$$
Step 3: Significant Figures for the answer: $$ \mathrm{q} = 26,000 ~(2 SF) \\ \mathrm{mass} = 630 ~(2 SF) \\ \mathrm{Specific~Heat} = 1.687 ~(4 SF)
$$
QUESTION

The following Practice is in TEST MODE and may not properly keep track of SF,
Take a screen capture/photo of any errors and report to Jay

Using the Specific Heat data on your Cheat Sheet answer the following question:
What is the change in Temperature (°C) of a sample of Ethyl Chloride, weighing 630 grams has $ 26,000 $ J of energy added?