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Standard X

Physics

Specific Heat Capacity

A copper bloc...

Question

A copper block of mass $2.5 kg$ is heated in a furnace to a temperature of $500^{\circ} C$ and then placed on large ice block. The maximum amount of ice that can melt is (Specific heat of copper $= 0.39 {Jg}^{- 1}^{\circ} C^{- 1}$ , latent heat of fusion of water $= 335 {Jg}^{- 1})$

1.2 kg

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1.455 kg

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1 kg

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2.5 kg

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Solution

The correct option is B $1.455 kg$
By the principle of calorimetry
Heat gained by ice = heat lost by copper block
$m_{1} L_{f} = m_{c} s_{c} Δ T_{c} \Rightarrow m_{1} = \frac{m_{c} S_{c} Δ T_{c}}{L - f} = \frac{2.5 \times 0.39 \times (500 - 0)}{335} = 1.455 kg$ .

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Similar questions

Q. A copper block of mass

2.5 kg

is heated in a furnance to a temperature of

500^{\circ} C

and then placed on a large ice block of temperature

0^{\circ} C

. The maximum amount of ice that can melt is:
(Specific heat of copper

= 0.39 J g^{- 1 \circ} C^{- 1}

, latent heat of fusion of ice

= 334 {Jg}^{- 1}

)

Q. A copper block of mass 2.5 kg is heated in a furnace to a temperature of

500^{o} C

and then placed on a large ice block. What is the maximum amount of ice that can melt ? (Specific heat of copper

= 0.39 J g^{- 1} K^{- 1}

; heat of fusion of water

= 335 J g^{- 1})

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block.Specific heat of copper =0.39Jg^-1K^-1,heat of fusion of water =335Jg^-1 Maximum amount of ice that can be melted

A)1 kg. B)1.5 kg. C) 2 kg. D)2.5 kg

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g^–1K^–1; heat of fusion of water = 335 J g^–1).

Q. A copper block of mass

2 kg

is heated to a temperature of

500^{\circ} C

and then placed in a large block of ice at

0^{\circ} C

. What is the maximum amount of ice that can melt? The specific heat of copper is

400 {J kg}^{- 1}

^{\circ} C^{- 1}

and latent heat of fusion of water is

3.5 \times 10^{5} {J kg}^{- 1}

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A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500∘ C and then placed on large ice block. The maximum amount of ice that can melt is (Specific heat of copper =0.39 Jg−1 ∘ C−1, latent heat of fusion of water =335 Jg–1)

The correct option is B 1.455 kgBy the principle of calorimetry Heat gained by ice = heat lost by copper block m1Lf=mcsc Δ Tc⇒m1=mcSc Δ TcL−f=2.5×0.39×(500−0)335=1.455 kg.

A copper block of mass $2.5 kg$ is heated in a furnace to a temperature of $500^{\circ} C$ and then placed on large ice block. The maximum amount of ice that can melt is (Specific heat of copper $= 0.39 {Jg}^{- 1}^{\circ} C^{- 1}$ , latent heat of fusion of water $= 335 {Jg}^{- 1})$

The correct option is B $1.455 kg$
By the principle of calorimetry
Heat gained by ice = heat lost by copper block
$m_{1} L_{f} = m_{c} s_{c} Δ T_{c} \Rightarrow m_{1} = \frac{m_{c} S_{c} Δ T_{c}}{L - f} = \frac{2.5 \times 0.39 \times (500 - 0)}{335} = 1.455 kg$ .