Use the Retreaces te ce In the laboratory Astuden t heats The heat capacity of t

Question

Use the Retreaces te ce In the laboratory Astuden t heats The heat capacity of the calorimeter (sometimes refetred to as the calorimeter constan) was determined in a separate Assoming that no beat as lost to the surroundings caleulate the specific heat of lead a"eoffee cup" calorimeter, or constant pressare calorimeter, is frequently used to determine the specific heat of a solid, or to measure the mergy of a solation ?hase reaction 61.21 grans oflead to 98.12 °? Rndthen drops it into a cup contaning S1.21gsams ofwater at 21.68 °CShe measureste final temperature to be 23.90 ? 8 more group attempts remaihing

Explanation / Answer

Ans. # Temperature change of water, dT1 = (final – initial) temp. = 23.900C – 21.680C

= 2.220C

# Mass of water = 81.21 g

# Temperature change of metal, dT2 = 23.900C – 98.120C = -74.220C

# Mass of metal = 61.21 g

# When hot metal is placed in contact with cold water, the hot metal loses some heat, and the water gains the same amount of heat in order to attain the thermal equilibrium.

Amount of heat gained/lost by the samples is given by-

q = m s dT                            - equation 1

Where,

q = heat gained/lost

m = mass of sample

s1 = specific heat of sample

dT = Final temperature – Initial temperature

Now,

            Heat lost by metal = Heat gained by water

            Or, -q (metal) = +q (water)             ; the –ve sign indicates heat loss

            Or, -[61.21 g x s x (-74.22)0C] = 81.21 g x 4.184 J g-10C-1 (2.22)0C

            Or, 4543.0062 s g 0C = 754.3174608 J

            Or, s = 754.3174608 J / 4543.0062 g 0C = 0.166 J g-10C-1

Therefore, specific heat of the lead = 0.166 J g-10C-1

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.