Explanation:
Here, Mass of water m
w
=100g
Mass of ice, m
i
=10g
Specific heat of water,S
=1calg
−1
o
C
Latent heat of fusion of ice,L
fi
=80calg
Let T be the final temperature of the mixture.
Amount of heat lost by water
=m
s
(△T)
=100×1×(50−T)
Amount of heat gained by ice
L
+m
=10×80+10×1×(T−0)
According to principle of calorimetry:
Heat lost = Heat gained
100×1×(50−T)=10×80+10×1×(T−0)
500−10T=80+T
11T=420orT=38.2
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Answers & Comments
Explanation:
Here, Mass of water m
w
=100g
Mass of ice, m
i
=10g
Specific heat of water,S
w
=1calg
−1
o
C
−1
Latent heat of fusion of ice,L
fi
=80calg
−1
Let T be the final temperature of the mixture.
Amount of heat lost by water
=m
w
s
w
(△T)
w
=100×1×(50−T)
Amount of heat gained by ice
=m
i
L
fi
+m
i
s
w
(△T)
i
=10×80+10×1×(T−0)
According to principle of calorimetry:
Heat lost = Heat gained
100×1×(50−T)=10×80+10×1×(T−0)
500−10T=80+T
11T=420orT=38.2