Answer:
5232.5J
Explanation:
Data obtained from the question. This includes the following:
Mass (M) = 250g
Initial temperature (T1) = 30°C
Final temperature (T2) = 25°C
Change in temperature (ΔT) = T1 – T2 = 30°C – 25°C = 5°C
Specific heat capacity (C) = 4.186J/g°C
Heat (Q) =.?
The heat released can be obtained as follow:
Q = MCΔT
Q = 250 x 4.186 x 5
Q = 5232.5J
Therefore, the heat released is 5232.5J
The amount of heat released when the bottle (250g) of water is cooled is 5232.5J .
Given data:-
Mass (M) = 250g
Initial temperature (T1) = 30°C
Final temperature (T2) = 25°C
Change in temperature (ΔT) = T1 – T2 = 30°C – 25°C = 5°C
Specific heat capacity (C) = 4.186J/g°C
Heat (Q) =.?
The heat released can be obtained as follow:
[tex]Q = MC\delta T\\\\Q = 250\times 4.186 \times 5\\\\Q = 5232.5J[/tex]
Therefore, the heat released is 5232.5J.
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