A piece of copper metal (specific heat 0.385J/g degree C ) at 86.5degree C is placed in 50.0g of water at 16.4 degree C. The metal and water come to the same temperature of 23.9 degreeC
First answer is correct. Note that the specific heat of copper is not needed in this computation.
Hello ssssss, your question comes under the category of physics, but still its a simple one. Here's how to solve it: We know that, In thermal equilibrium, Heat lost by a hot body equals heat gained by a cold body. Here, the hot body is the piece of copper, since is temperature is higher as compared to water. The common temperature 't' attained by both is 23.9 degrees C. So, Let m(copper) be the mass of copper. So we have the following parameters: t = 23.9 degrees C m(copper) = ? c(copper) = .385 J/g degrees C t(copper) = 86.5 degrees C m(water) = 50g c(water) = 4.2 J/g degrees C t(water) = 16.4 degrees C So, heat lost by hot copper metal = heat gained by cold water. m(copper) x c(copper) x ( t(copper) - t ) = m(water) x c(water) x ( t - t(water) ) m(copper) x 24.101 = 1575 Or, m(copper) = 1575/24.101 m(copper) = 65.34999 g = 65.35 g Wasn't it simple! Hope my steps were clear to you By W@W.
