Question
Download Solution PDFA 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. The maximum amount of ice that can be melt, will be :
Given :
Specific heat of copper = 0.4 Jg−1 °C−1
Heat of fusion of water = 400 Jg−1
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
When copper block after getting heated by a furnace is kept in a block of ice, the heat is liberated by copper block and recieved by block of ice i.e.
Heat lost by copper block = Heat gain by ice
Calculation:
Given:
mc = mass of copper block = 2.5 kg, tc = 500°C, Ccopper = 0.4 Jg−1°C−1
L = heat of fusion of water = 400 Jg−1
Heat lost by copper block = Heat gain by ice
mcCcoppertc = miceL
2.5 × 0.4 × 500 = mice × 400
mice = 1.25 kg
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