A 100g block of metal with a specific heat of 0.2 cal/gC that is at 100C is drop

Question

A 100g block of metal with a specific heat of 0.2 cal/gC that is at 100C is dropped into a large Styrofoam cup that contains 100g of water that is at 20C. A) Compare the heat lost/gained by the water and the metal block. A numerical value is not needed here, but explain your response. B) Compare the temperature change of the water and the metal block. Which will change temperature the most? Numerical values are not needed at this point, but explain your response. C) What is the final temperature of the mixture? Do a calculation and start with energy conservation: Qh+Qc=0 THANKS!!

Explanation / Answer

heat loss by the block=heat gain by water m(block)*c(block)*delta(T,block)=m(water)*c(water)*delta(T,water) change in temperature of block is more because its specific heat is less 100*0.2*4.18*(100-T)=100*4.187*(T-20) 8360-83.6*T=418.7*T-8374 T=33.31 degree c final temp=33.31

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