Question

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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 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 ).

Answer 1

Given :

Mass of the copper block ( m ) = 2.5 kg .

We know that 1 kg = 1000 g

Hence 2.5 kg = 2.5 × 1000 g = 2500 g .

Specific heat capacity ( C ) = 0.39 J g⁻¹ .

Rise in temperature of the copper block ( Δt ) = 500°C = 500 K

We know that the amount of heat = m C Δt .

Q = m C Δt

⇒ Q = 2500 g × 0.39 J K/g × 500 K

⇒ Q = 975 × 500 J

⇒ Q = 487500 J

Let the maximum amount of ice that it can melt be m .

According to the specific latent of heat :

Q = m L where L is specific latent heat .

Q = 487500 J ( calculated before ) .

L = 335 J/g ( as given )

Q = m L

⇒ m = Q/L

⇒ m = 487500 J / 335 J/g

⇒ m = 487500 / 335 g

⇒ m = 1455.22 g

The maximum amount of ice that it can melt is 1455.22 g .

Answer 2

ANSWER:------------

EXPLANATION:-

{maximum heat loss by copper block =mSdT}

we know that,

{=2500g x 0.39jg-1k-1 x (500-0) }

[=487,500 j ]

now ,

{487500 j=M L}

[NOTE]=

{ M is the mass of ice melted}

{ L is latent heat of fusion }

{487500 j =M x 335jg-1}

{M=1455.22 g =1.45522 kg}

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