A piece of iron is heated to 95.0 celsius and then placed in an insulated vessel containing 250. grams of water at 25.0 celsius. when the system comes to equilibrium the temperature of the system is 35.0 celsius. what is the mass in g of the iron? assume no heat is lost to the surroundingsSpecific heat/J. g^-1 K^-1iron 0.450water 4.184 (A) 4.48 g (B) 41.7 g (C) 387 g (D) 612 g

The mass of the iron has been calculated as 387.40 grams.

Specific heat can be defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree celsius.

Since in the reaction there has been no heat loss, the heat released by iron has been equivalent to heat absorbed by the water in the calorimeter.

Heat = mass × specific heat of water × change in temperature

Heat absorbed = 250 grams × 4.184 J/g × (308 K - 298 K)

Heat absorbed = 10460 J

Heat released = mass × specific heat of water × change in temperature

Heat released = mass × 0.45 J/g.K × (308 K - 368 K)

Heat released = mass × 27

Heat released = Heat absorbed

Mass of Iron × 27 = 10460 J

Mass of iron = 387.40 grams.

The mass of the iron has been calculated as 387.40 grams.

For more information about specific heat, refer to the link:

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c) 387g

Explanation:

Water;

Mass = 250g

Specific heat = 4.184

Initial Temp, T1 = 25 + 273 = 298K

Final Temp, T2 = 35 + 273 = 308K

Heat = ?

H = mc(T2 - T1)

H = 250 * 4.184 (308 - 298)

H = 10460 J

Iron;

Initial Temp, T2 = 95 + 273 = 368K (Upon converting to kelvin temperature)

Mass = ?

Final Temp, T1 = 35 + 273 = 308

Heat = 10460 (Heat lost by iron is qual to heat gained by water)

Specific heat = 0.45

H = mc(T2-T1)

M = 10460 / [0.45 (308 - 368)]

M = 10460 / 27

M = 387g

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