What quantity of heat would be given out by 200 gm of copper in cooling from 80 -C to 20 -C Specific heat of copper - 0-09 cal g -1 - C -1-

The correct option is A 1080 cal Given , #160; m=200g , #160;θ_1=80^0C , θ_2=20^0C We have , specific heat of copper c=0.09cal/g ^oC Now , ...

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Byju's Answer

Standard X

Physics

Specific Heat Capacity

What quantity...

Question

What quantity of heat would be given out by $200 g m$ of copper in cooling from $80^{\circ} C$ to $20^{\circ} C$ (Specific heat of copper = 0.09 ${c a l g}^{- 1}$ $^{\circ} C^{- 1}$ )?

1080 c a l

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1000 c a l

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1500 c a l

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1100 c a l

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Solution

The correct option is A $1080 c a l$
Given , $m = 200 g, θ_{1} = 80^{0} C, θ_{2} = 20^{0} C$
We have , specific heat of copper $c = 0.09 c a l / g -^{o} C$
Now , heat lost by copper is given by the definition of specific heat c ,
$Q = m c Δ θ = m c (θ_{1} - θ_{2})$
or $Q = 200 \times 0.09 \times (80 - 20) = 1080 c a l$

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Similar questions

$50 g$ of copper is heated to increase its temperature by $10^{\circ} C$ . If the same quantity of heat is given to $10 g$ of water, then what would be the rise in its temperature?
(specific heat of copper $= 420 J / k g /^{\circ} C$ , specific heat of water $= 4200 J / k g /^{\circ} C$ )

Q. Find the heat lost by a copper cube of mass

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$50 g$ of copper is heated to increase its temperature by $10^{\circ} C .$ If the same quantity of heat is given to $10 g$ of water, the rise in its temperature is:

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What quantity of heat would be given out by 200gm of copper in cooling from 80∘C to 20∘C (Specific heat of copper = 0.09 calg−1∘C−1)?

The correct option is A 1080calGiven , m=200g,θ1=800C,θ2=200CWe have , specific heat of copper c=0.09cal/g−oCNow , heat lost by copper is given by the definition of specific heat c , Q=mcΔθ=mc(θ1−θ2)or Q=200×0.09×(80−20)=1080cal

What quantity of heat would be given out by $200 g m$ of copper in cooling from $80^{\circ} C$ to $20^{\circ} C$ (Specific heat of copper = 0.09 ${c a l g}^{- 1}$ $^{\circ} C^{- 1}$ )?