Given:

A Copper Block Of

Mass of the copper block = 2.5 Kg = 2500g

Temperature = [latex] 500^{\circ}C [/latex]

Specific Heat Of Copper =[latex] 0.39 J/g^{-1} C [/latex]

Heat Of Fusion Of Water =[latex] 335 J/g^{-1} [/latex]

The maximum heat the copper block can lose is

[latex] \Rightarrow Q = mC\Delta \Theta [/latex] [latex] \Rightarrow 2500 * 0.39 * 500 [/latex] [latex] \Rightarrow 487500J [/latex]

Let[latex] m_{1}g [/latex] be the amount of ice that melts when the copper block is placed on the ice block.

The heat gained by the melted ice,[latex] Q = m_{1}L [/latex]

Therefore, [latex] m_{1} = \frac{Q}{L} [/latex] [latex] \Rightarrow m_{1} = \frac{487500}{335} [/latex] [latex] \Rightarrow m_{1} = 1455.22g [/latex]

Therefore, the maximum amount of ice that can melt is 1.45 kg.

Explore more such questions and answers at BYJU’S.

Was this answer helpful?

  
   

0 (0)

Upvote (0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question