It is given that, Mass (m) = 5.00kg

Temperature (ΔT) = 80.0 K

Heat supplied (Q)= 1676 KJ= 16,76,000 J

Specific heat can be calculated by the formula,

C = \frac{Q}{m \Delta T}

Substituting the known values in the above formula, we get,

C = \frac{16,76,000 J}{(5kg)(80 K)}

C = 4190 J kg-1 K -1

Hence, the specific heat of water is 4190 J kg-1 K -1 .

I found an answer from ntrs.nasa.gov

General Disclaimer One or more of the Following Statements may ...

broad **heat transfer** data base to **determine** where **it** could appropriately be used for hrat ... (5) The **hot water** loop: **heated** the preplate and recovery walls, using ...

For more information, see General Disclaimer One or more of the Following Statements may ...

I found an answer from www.quora.com

Will cold **water** boil faster than **hot water**? - Quora

You need to add a lot of energy to liquid **water** in order to boil **it**. ... Um, call me dumb but I don't **see** how, all else being equal, cold **water** could ever ... **when** **heat energy** is added to a substance, the **temperature** will change by a certain amount. ... of the **pot**, the environment conditions, rate of **heat transfer**, the method of **heat** ...

For more information, see Will cold **water** boil faster than **hot water**? - Quora