Test Index

Thermal Properties of Matter

© examsnet.com
Question : 14 of 22
Marks: +1, -0
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Solution:  
Here, mass of metal, m=0.20 kg=200 gm = 0.20\ \mathrm{kg} = 200\ \mathrm{g}
Fall in temperature of metal
ΔT=15040=110C\Delta T = 150 - 40 = 110^{\circ}\mathrm{C}
If c is specific heat of the metal, then heat lost by the metal
,ΔQ=mcDT=200×c×110\Delta Q = mcDT = 200 \times c \times 110
Volume of water =150 c.c.= 150\ \mathrm{c.c.}
Mass of water, m=150 g\therefore \text{Mass of water, } m' = 150\ \mathrm{g}
Water equivalent of calorimeter, w=0.025 kg=25 gw = 0.025\ \mathrm{kg} = 25\ \mathrm{g}
Rise in temperature of water and calorimeter, ΔT=4027=13C\Delta T' = 40 - 27 = 13^{\circ}\mathrm{C}
Heat gained by water and calorimeter,
ΔQ=(m+w)ΔT\Delta Q' = (m' + w)\Delta T'
=(150+25)×13=175×13= (150 + 25) \times 13 = 175 \times 13
As, ΔQ=ΔQ\text{As, } \Delta Q = \Delta Q'
∴ From (i) and (ii),200×c×110=175×13200 \times c \times 110 = 175 \times 13
or c=175×13200×1100.1c = \frac{175 \times 13}{200 \times 110} \approx 0.1
If some heat is lost to the surroundings, value of c so obtained will be less than the actual value of c.c.
© examsnet.com
Go to Question: