What about:

In[47]:= NonlinearModelFit[data, {temp2[t], a <= 0 || a > 0}, {a, k2},  t]["BestFitParameters"]
Out[47]= {a -> 866.073, k2 -> 0.0300234}

Edit:

In[30]:= NonlinearModelFit[data, {temp2[t], 1 > 0}, {a, k2}, t]["BestFitParameters"]
Out[30]= {a -> 866.073, k2 -> 0.0300234}
In[31]:= NonlinearModelFit[data, {temp2[t], True}, {a, k2}, t]["BestFitParameters"]
Out[31]= {a -> 866.073, k2 -> 0.0300234}
In[32]:= NonlinearModelFit[data, {temp2[t]}, {a, k2}, t]["BestFitParameters"]
Out[32]= {a -> 339.024, k2 -> 9129.04}

Thanks. I should have tried starting values. So it was only a lucky accident that I found a good fit in all cases, using different data sets, were also k1 was a free parameter?

There are 2 issues:

(1) FindFit should be avoided as it doesn't have any capabilities to provide diagnostic measures (such as standard errors, confidence bands, residuals, etc.). NonlinearModelFit should be used instead.

(2) You need better starting values than the defaults that FindFit uses (although this applies also to NonlinearModelFit). Consider the following:

temp2[t_] := a*(Exp[-0.01376208490662561*t] - Exp[-k2*t])
FindFit[data, temp2[t], {{a, 860}, {k2, 0.03}}, t]
(* {a -> 866.073, k2 -> 0.0300234} *)

You get exactly what you had before. Good starting values are your best friends in an iterative procedure.