42.5 grams of an unknown substance is heated to 105.0 degrees celsius and then placed into a calorimeter containing 110.0 grams of water at 24.2 degrees celsius. if the final temperature reached in the calorimeter is 32.4 degrees celsius, what is the specific heat of the unknown substance?
show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 j/(° c × g).

The specific heat of the unknown substance is 1.22 J/g.°C.

Explanation:

Heat lost by substance (Qc) = Heat gained by the water (Qw) ,

- (Qc) = (Qw).

Qw = m.c.ΔT,

where, Qw is the amount of heat released to water (Q = ??? J).

m is the mass of water (m = 110.0 g).

c is the specific heat capacity of solution (c = 4.18 J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = 32.4°C - 24.2°C = 8.2°C).

∴ Q = m.c.ΔT = (110.0 g)(4.18 J/g.°C)(8.2°C) = 3770.36 J.

(Qc) = m.c.ΔT,

where, Qc is the amount of heat lost by substance (Qc = - 3770.36 J).

m is the mass of water (m = 42.5 g).

c is the specific heat capacity of solution (c = ??? J/g.°C).

ΔT is the difference in T (ΔT = final temperature - initial temperature = 32.4°C - 105.0°C = -72.6°C).

∴ (- 3770.36 J) = (42.5 g)(c)(-72.6°C).

∴ c = (- 3770.36 J)/(42.5 g)(-72.6°C) = 1.222 J/g.°C.

under a microscope, cancer cells may look very different from normal cells. they often have different sizes and some may be larger than normal while others are smaller. cancer cells are often abnormally shaped and the control centre of the cell (the nucleus) may have an abnormal appearance.

i think? ?