11.3WTF2 1.1 x 10^{1}
WTF3 11
Step 1: Solve using the equation: $$ \Delta T = \dfrac{q}{m \cdot s} $$
Step 2: Using a bar problem: $$ \dfrac{39,000~J}{} \times \dfrac{1}{830 ~g} \times \dfrac{g^{\circ}C}{4.184 ~J} = {11 ~^{\circ} C \atop ~or~ 1.1 x 10^{1} ~^{\circ} C }$$
Step 3: Significant Figures for the answer: $$ \mathrm{q} = 39,000 ~(2 SF) \\ \mathrm{mass} = 830 ~(2 SF) \\ \mathrm{Specific~Heat} = 4.184 ~(4 SF)
$$
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 H2O, weighing 830 grams has $ 39,000 $ J of energy added?