It is. How does it relate to the cooling of the metal?lxhull said:Not sure if its relevant
That's the problem, I don't know. It seems like it can't be part of the equation for the metal's shc because it used the waters shc, so I can't figure it out.haruspex said:It is. How does it relate to the cooling of the metal?
Just write the corresponding equation for the metal's change in temperature. Create an unknown for the metal's s.h.lxhull said:That's the problem, I don't know. It seems like it can't be part of the equation for the metal's shc because it used the waters shc, so I can't figure it out.
Corrrct so far.lxhull said:Homework Statement:: A thermos bottle contains 0.150 kg of water at 4.1 °C. When 9.00 x 10^-2 kg of a metal, initially at 96.2 °C, is put into the water, the temperature of the water rises to 21.7 °C. Calculate the specific heat of the metal
Relevant Equations:: C= Eth/mT
Previously solved thermal energy gained by water as
Eth= 0.15(4180)(17.6) = 11035.2 J
Not sure if its relevant