Should I use a mixed-effects model?. Measures at different locations in several pieces Imagine a simple experiment:
I have 10 "almost identical" pieces.
I take the first one and I measure the temperature at 5 different distances from its center.
I take the second one and I do the same 1 hour later.
I take the third one and I do the same 2 hour laters.
and so on till the last piece (I measure its temperatures at 5 different positions 9 hours later).   
My teacher has donde a simple regression model like (R syntax)
Temp  ~ Time + Position

But, is it right?
From my point of view the measures are not independent.
Shouldn't we use a mixed effects model like this?
Temp  ~ Time + Position + (1|Piece)

The problem I see here is that in such a model the temperature variation could be wrongly attributed to being a different piece instead of being due to the time.
PD:In fact there are not just one piece per time, it could be up to 3, but I've simplified the problem. 
REPLIES:
The same piece is never measured at two time points. True.
The research question is:  How the temperature varies along the position and time (in a linear model).
The pieces start at the same temperature at time zero, and their temperature varies along time, I don't know how, I need to extract that information from the measures.
They cool without independently, without contact, but are supposed to cool on the same way.
In fact my model is not exactly this one but I've written the question like this because it's equivalent and I thought it could be easier.
In the real model I want to measure how different graftings degrade through time.  And I quantify it measuring their absortivity with a CT scan, taking readings at different positions on each grafting.
A simplified graphic:

Each ball is measured once at different times.
Each ball is measured at several positions.
 A: I agree that the measures are not independent because each measurement is nested within a piece and there are several measurements per piece.
The 5 measurements on piece A may be more similar to each other than the 5 measurements on piece B (other than the difference in the time they were taken). This might be because of physical differences between the pieces: after all, they are "almost" identical and not "absolutely" identical. Or, possibly something concerning the experiment is different at the times each piece is measured, for instance if the measurements are not automated and taken simultaneously, there will be some interval between the measurements on each piece, which could be affected by the experimenter (what he/she drank or ate between the measurements and if the measurements were taken by the same person at each time. This is speculation obviously, the point is that there /could/ be non-independence, so it seems very prudent to allow for any non-independence by including random intercepts for Piece as you suggest. A mixed model will estimate the variance of a random intercept for the pieces while still estimating the fixed effects of Time and Position. 
It seems very prudent to allow for any non-independence by including random intercepts for Piece as you suggest. A mixed model will estimate the variance of a random intercept for the pieces, while still estimating the fixed effects of Time and Position. 
Edit: a simulation to demonstrate that a random intercepts model should work
We define a data generating process of
$$ y = 300 - 10t - 20p$$
where $t$ represents time, and $p$ represents the position of measurement, so that the coefficients for time and position should be $-10$ and $-20$ respectively.
In the original post, $t \in (1,5)$ and $p \in (1,10)$
So we simulate data for N = 10 pieces, with n_per_piece = 5 measurements each, as per the original question. 
require(lme4)
set.seed(987)
# number of pieces / times
N <- 10

# Random variation between pieces
# (that is, a random intercept for pieces)
ran_int <- rnorm(N,0,3)  

# measurements per piece (positions)
n_per_piece <- 5

# True mean at each measurement occasion for each position
# data generating process:
# measurement = 300 - time*10 - position*20
dtg <- function (t,p) return(300 - t*10 - p*20)

dt <- data.frame()

for (i in 1:N) {
  # loop over pieces ()

  # mean measurements for piece i
  mu_vec = dtg(i,c(1:n_per_piece))

  measures <- rnorm(n_per_piece, mu_vec+ran_int[i], 7) 

  piece <- rep(i,n_per_piece)
  position <- 1:n_per_piece

  dt <- rbind(dt,cbind.data.frame(piece,position,measures))
}

dt$time <- dt$piece
dt$piece <- as.factor(dt$piece)

m0 <- lmer(measures~position+time+(1|piece),data=dt)
summary(m0)

Note that the standard deviation of the random effect was specified as 3, and the residual standard deviation as 7.
And we see from the model output:
Random effects:
 Groups   Name        Variance Std.Dev.
 piece    (Intercept) 10.66    3.266   
 Residual             48.74    6.982   
Number of obs: 50, groups:  piece, 10

Fixed effects:
            Estimate Std. Error t value
(Intercept) 302.3960     3.7300   81.07
position    -20.2144     0.6982  -28.95
time        -10.2024     0.4974  -20.51

which agrees with the parameters we used to generate the data. Specifically, the fixed effects are estimated as $-20$ and -$10$, the random intercept standard deviation is $3.3$ and the residual standard deviation is $6.9$
