Q. In an experiment, 100 g of water requires 12 600 J of heat to
raise it from 30 ⁰C to 60 ⁰C.
a) Find the heat capacity of 100 g of water.
b) Find the heat capacity of 1000 g of water.
Related to Thermal Properties of Matter......
Please help me!
Answers
Answer:
Given: Specific heat capacity of copper =0.4Jg
−1
o
C
−1
Specific heat capacity of water =4.2Jg
−1
o
C
−1
Specific latent heat of fusion of ice =336Jg
−1
Heat given by water to reach 5
o
C+ Heat given by copper vessel to reach 5
o
C= Heat taken by ice to melt at 0
o
C+ Heat taken by melted ice to reach 5
o
C
m
w
c
w
Δt+m
c
c
c
Δt=m
i
L+m
i
c
w
δt
where Δt= change in temperature of water and copper vessel
m
w
= mass of water = 150g
m
c
= mass of copper vessel = 100g
m
i
= mass of ice
c
c
= specific heat capacity of copper
c
w
= specific heat capacity of water
L = specific latent heat of fusion of ice
δt = change in temperature of ice
Therefore, [150×4.2×(50−5)]+[100×0.4×(50−5)]=(m×336)+[m×4.2×(5−0)]
(150×4.2×45)+(100×0.4×45)=(m×336)+(m×4.2×m)
28350+1800=336m+21m
357m=30150
m=
357
30150
=84.45g of ice
Hence, 84.45g of ice is needed to cool it to 5°C
Mark as brainliest