Question

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°cand then placed on large ice block. what is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J/g C and the latent Heat of
fusion of water = 335 J/g)​

Answer 1

Answer:

we know

heat loss =heat gain

maximum heat loss by copper block =mSdT

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

=487,500 j

now

487500 j=M L

where M is the mass of ice melted and L is latent heat of fusion .

487500 j =M x 335jg-1

M=1455.22 g =1.45522 kg

Answer 2

Answer:

• $1455.22 g$

Explanation:

• Mass of the copper block, $m = 2.5 kg = 2500 g$
• The Rise in the temperature of the copper block, $\Delta \theta = 500^{\circ}C$
• Specific heat of copper, $C = 0.39Jg^{-1^{o}} C^{-1}$
• Maximum heat the copper block can lose, $Q = mC\Delta \theta$

$\implies\:\:\: 2500 \times 0.39 \times 500$

$\implies\:\:\:487500J$

Let,

• $m_{1}g$ be the amount of ice that melts when the copper block is placed on the ice block.

Hence, The heat gained by the melted ice, $Q = m_{1}L$

$\therefore \:m_{1} = \dfrac{Q}{L}$

$\implies\:\:\: \dfrac{487500}{335}$

$\implies\:\:\:1455.22 g$

Hence,

• Maximum amount of ice that can melt is 1.45 kg. (1455.22 g)