Question

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

Answer 1

Answer:

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