50g of a certain metal at 150∘C is immersed in 100g of water at 11∘C. The final temperature is 20∘C. Find the specific heat capacity of the metal. Specific heat capacity of water is 4.2Jg−1K−1).
0.582
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Solution
The correct option is A 0.582 Given: Mass of metal, mm=50g Mass of water, mw=100g Temperature of metal =150∘C Temperature of water =11∘C Temperature of the mixture =20∘C Specific heat of water, Sw=4.2Jg−1K−1
Heat lost by the metal = Heat gained by water mmsmΔTm=−mwswΔTw 50×sm×(150−20)=−100×4.2×(11−20 sm=0.582Jg−1K−1