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Standard VII

Physics

Effects of Heat

Find out the ...

Question

Find out the increase in temperature of $50 g$ block of solid marble (specific heat $= 0.9 k J / k g$ ), originally at $20$ if it absorbs $100 J$ of heat ?

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Solution

The correct option is B $2$
We know that heat, $Q = m s Δ T$
or $Δ T = Q / m s = 100 / (50 \times 0.9) = 2.2 \sim 2$

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Similar questions

Q. 50g of copper is heated to increase its temperature by

10^{o} C

. If the same quantity of heat is given to 10 g of water, the rise in its temperature is :
(Specific heat of copper

=

420J/kg

^{o} C

; Specific heat of water

=

4200J/kg

^{o} C

)

A hot solid of mass 60 g at 100 $^{\circ} C$ is placed in 150 g of water at 20 $^{\circ} C$ . The final temperature recorded is 25 $^{\circ} C$ . Calculate the specific heat of solid. (Specific heat of water = 4200J/kg/ $^{\circ} C$ )

$50 g$ 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:

(Given: ${Specific heat of copper = 420 Joule kg}^{- 1}^{\circ} C^{- 1}$ , ${Specific heat of water = 4200 Joule kg}^{- 1}^{\circ} C^{- 1}$ )

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 (specific heat of copper

= 420 J / k g /^{\circ} C

)

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 the rise in its temperature is (specific heat of copper

= 420 J / k g /^{\circ} C

)

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Find out the increase in temperature of 50g block of solid marble (specific heat =0.9kJ/kg), originally at 20 if it absorbs 100Jof heat ?

The correct option is B 2We know that heat, Q=msΔTor ΔT=Q/ms=100/(50×0.9)=2.2∼2

Find out the increase in temperature of $50 g$ block of solid marble (specific heat $= 0.9 k J / k g$ ), originally at $20$ if it absorbs $100 J$ of heat ?

The correct option is B $2$
We know that heat, $Q = m s Δ T$
or $Δ T = Q / m s = 100 / (50 \times 0.9) = 2.2 \sim 2$