500 g of hot water at 60 -C is kept in the open till its temperature falls to 40 -C- Calculate the heat energy lost to the surroundings by the water- specific heat of water - 4200 J kg -1 - C -1

Calculate the height of a water-fall through which the fall of water leads to increase in its temperature by 1⁰C. Assume no energy is lost during process.

Specific heat capacity of water = $4200 J / k g^{\circ} C$

Accelaration due to gravity = $10 m / s^{2}$

The amount of heat energy required to convert 1 kg of ice at $- 10^{\circ} C$ to water at $100^{\circ} C$ is 7,77,000 J. Calculate the specific latent heat of ice.

Specific heat capacity of ice = 2100 J $k g^{- 1} K^{- 1}$ , specific heat capacity of water=4200 J k $g^{- 1} K^{- 1}$ .

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500g of hot water at 60∘C is kept in the open till its temperature falls to 40∘C. Calculate the heat energy lost to the surroundings by the water. ( specific heat of water = 4200 J kg−1∘C−1)

The correct option is D 42000JGiven , m=500g=0.5kg,θ1=600C,θ2=400CWe have , specific heat of water c=4200J/kg−oCNow , heat lost to the surroundings by the water is given by the definition of specific heat c , Q=mcΔθ=mc(θ1−θ2)or Q=0.5×4200×(60−40)=42000J

$500 g$ of hot water at $60^{\circ} C$ is kept in the open till its temperature falls to $40^{\circ} C$ . Calculate the heat energy lost to the surroundings by the water. ( specific heat of water = 4200 J ${k g}^{- 1}^{\circ} C^{- 1}$ )