Question

5. 500 g of water at 60°C is kept in the open till its temperature
drops to 50°C. Calculate the heat lost to the surroundings. The
specific heat of water is 4200 J/kg°C.​

Answer 1

Explanation:

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Answer 2

hlo mate here is ur ans........

m=500,g=0.5kg

ΔT=60−40=20oC

C=4200Jkg^−1C^−1Q=?

Using the formula, Q=mCΔT

=0.5×4200×20=42000J

∴Heat lost=42,000J

hope this helps uh dear :)

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