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ahrayia [7]
3 years ago
13

How much heat is released when 432 g of water cools down from 71'c to 18'c?

Physics
1 answer:
maria [59]3 years ago
7 0
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
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