It suffices to suppose that one of your complexes, say $K$, is an acyclic complex of free abelian groups. Then, if $L$ is any complex of abelian groups, $K \otimes L$ is acyclic. 

This can be easily seen as follows: Assume for the moment we know that $K$ is contractible, i.e. there is a chain equivalence $f:  K \to 0$. Then $f \otimes id_L:K \otimes L \to 0 \otimes L=0$ is a also a chain equivalence (see MacLane, Homotopy, Cor. V.9.2). Hence $H_\ast(K \otimes L) \cong H_\ast(0)=0$ shows that $K \otimes L$ is acyclic. 

The fact that acyclic complexes of free abelian groups are contractible is Lemma VI.3.2 of Brown: Cohomology of Groups. But the proof is simple enough to be posted here. First note 
that the short exact sequence 
$$0 \to \operatorname{im} d_{i+1} \to K_i \to \operatorname{im} d_i \to 0$$
splits since as a subgroup of the free abelian group $K_{i-1}$, $\operatorname{im} d_i$ is a free abelian group and hence a projective $\mathbb Z$-module. Write 
$$K_i = \operatorname{im} d_{i+1} \oplus A_i$$
such that $d_i$ maps $A_i$ isomorphicaly onto $\operatorname{im} d_i$. Let $e_i: \operatorname{im} d_i \to A_i$ be the inverse map. Now define $h_i:K_i \to K_{i+1}$ by 
$$h_i|\operatorname{im} d_{i+1} := e_{i+1},\;\;\;h_i|A_i := 0$$
Let $x=d+a \in K_i$ with $d \in \operatorname{im} d_{i+1}, a \in A_i$. Since $d_i(x)=d_i(a)$ we obtain  
$$(h_{i-1}d_i + d_{i+1}h_i)(x) = e_i(d_i(a)) + d_{i+1}(e_i(d))= a + d=x.$$
Hence $h: id_K \simeq 0$ is the searched homotopy. 

**Added:** The argument holds verbatim for complexes over a PID.