I need help for solving specific heat for my upcoming exam. How can i compute the specific heat? Is there an easy technique? if there is, please tell me how?
Sample problem:
A thermos bottle contains 80g of water at 15 degree celcius. Into this placed 40g of metal at 85 degree celcius after equilibrium is established, the temparature of the water and metal is 35 degree celcius. What is the specific heat of metal? Assume no heat loss to the thermos bottle?
(if there is an answer of (3400gdegree celcius - (subtract) 1400g degree celcius) where do 1400g degree celcius comes from?
2007-02-23 13:03:11 · 2 answers · asked by Jeniv the Brit 7 in Science & Mathematics ➔ Physics
Key is to say "heat lost by hot" = "heat gained by cold"
(ms(Tinitial - Tfinal)) for hot = ms(Tfinal - Tinitial) for cold
80(1)(35-15) = 40(s)(85 -35) solve for s in cal/gdegC
Don't understand last bit
2007-02-23 13:15:39 · answer #1 · answered by hello 6 · 0⤊ 0⤋
All i understand is whilst i'm heating lead i do no longer choose as lots warmth as whilst i'm heating and melting aluminum so i will assert extra warmth is had to soften with the help of a blast fan into the coals hence elevating the temperature to 3 million,000 c quite of 350. c
2016-12-14 04:18:19 · answer #2 · answered by libbie 4 · 0⤊ 0⤋