Certain amount of heat is given to 100g of copper to increase its temperature by $21^{0} C$ . If the same amount of heat is given to 50 g of water, then what is the rise its temperature ?(Specific heat capacity of copper = 400 J / kgK and that for water = 42000 J / kgK ).

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Solution

The amount of heat supplied is given by
$Q = m c Δ T$
Here , $m = 100 g = 0.1 k g; c = 400 J / k g - K$
Thus, $Q = 0.1 \times 400 \times 21 = 840 J$
Hence, $840 = 0.05 \times 4200 \times Δ T$
$Δ T = 4^{o} C$

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Similar questions

Q. certain amount of heat is given to
100g of copper to increase its temperature by 21 deg c. if the same amount of heat is given to 50 g of water, then the rise in its temperature is (specific heat of copper = 400 j kg^-1 k ^-1and that for water is 4200 j kg^-1 k^-1)

50

g of copper is heated to increase its temperature by

10^{o}

C. If the same quantity of heat is given to

10

gm of water, the rise in temperature is (specific heat of copper

= 420

k g^{- 1} K^{- 1}

, specific heat of water

= 4200

k g^{- 1} K^{- 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, then 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, the rise in its temperature is (specific heat of copper = $420 J / k g /^{\circ} C$ , specific heat of water = $4200 J / k g /^{\circ} C$ )

Q. 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 is

420 J k g^{- 1}^{\circ} C

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Certain amount of heat is given to 100g of copper to increase its temperature by 210C. If the same amount of heat is given to 50 g of water, then what is the rise its temperature ?(Specific heat capacity of copper = 400 J / kgK and that for water = 42000 J / kgK ).

The amount of heat supplied is given by Q=mcΔTHere ,m=100g=0.1kg;c=400J/kg−KThus,Q=0.1×400×21=840JHence,840=0.05×4200×ΔTΔT=4oC

Certain amount of heat is given to 100g of copper to increase its temperature by $21^{0} C$ . If the same amount of heat is given to 50 g of water, then what is the rise its temperature ?(Specific heat capacity of copper = 400 J / kgK and that for water = 42000 J / kgK ).

The amount of heat supplied is given by
$Q = m c Δ T$
Here , $m = 100 g = 0.1 k g; c = 400 J / k g - K$
Thus, $Q = 0.1 \times 400 \times 21 = 840 J$
Hence, $840 = 0.05 \times 4200 \times Δ T$
$Δ T = 4^{o} C$