Apiece of iron (csfe = 0.45 j/g·⁰c) with a mass of 4.45 g at temperature of 101.0 oc is dropped into an insulated calorimeter containing 50.0 g of water at 22.00 oc. what will be the final temperature of the water?

22.75 ⁰C.

Explanation:

Q = m.c.ΔT.

where, Q is the amount of heat added,

m is the mass of the substance,

c is the specific heat of the substance,

ΔT is the temperature difference (final T - initial T).

∵ The amount of heat released by Fe = the amount of heat absorbed by water

∴ - (m.c.ΔT) released by Fe = (m.c.ΔT) absorbed by water.

- (4.45 g)(0.45 J/g·⁰C)(final T - 101.0 ⁰C) = (50.0 g)(4.186 J/g·⁰C)(final T - 22.0 ⁰C)

- 2.0025 final T + 202.2525 = 209.3 final T - 4604.

209.3 final T + 2.0025 final T = 202.2525 + 4604

211.3025 final T = 4806.2525.

final T = 4806.2525/211.3025 = 22.75 ⁰C.

what does the table look like