Water (2850 g ) is heated until it just begins to boil. if the water absorbs 5.53×105 j of heat in the process, what was the initial temperature of the water?

Answer is 54 °C.

Explanation;

We can simply use heat equation

Q = mcΔT  

 Where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Let's assume that the initial temperature is T.

Q = 5.53 × 10⁵ J

m = 2850 g

c = 4.186 J/g °C

ΔT = (100 - T)  °C     Since the water is boiling, the final temperature is 100 °C.

By applying the equation,

5.53 × 10⁵ J  = 2850 g  x 4.186 J/g °C  x (100 - T)  °C

(100 - T)  °C = 5.53 × 10⁵ J  / (2850 g  x 4.186 J/g °C  )

(100 - T)  °C = 46.35 °C

         T        = 100 - 46.35 C = 53.65 °C

                   ≈ 54 °C

catya that's not even a b or c

here is table 2-2:

country

percent of eligible voters that voted

rank

united states

44.9%

138

canada

60.1%

107

costa rica

84.6%

14

hungary

66.9%

82

japan

57.0%

118

egypt

27.7%

159

south africa

85.5%

10

ans is no, the data set cannot be plotted as a single circle graph