Calculating sum of heat transfered

I have a formula of thermal conductance heat transfer rate. Here it is: $$Q = \lambda{S (T_1 - T_2) \over L} \Delta t$$ For my calculations I have got some constant values available $$Q = 0.58{1 (T_1 - T_2) \over 0.01} \Delta t$$ I can express difference of $T_1$(90°C) and $T_2$(0°C) as (note: $c_1$ and $c_2$, $m_1$ and $m_2$ are equal for my values, so I join them) $$T_1 - T_2 = {Q_1\over c_1m_1} - {Q_2\over c_2m_2} = {Q_1 - Q_2 \over c_1m_1}$$ $$90 - 0 = {172431270 - 0\over2090\times 916,7}$$ for beginning values.

Now the problem: $T_1 - T_2$ is not constant, because $Q$ is just a rate of energy in one moment. In next moment, the difference of $T_1$ and $T_2$ will be smaller by $Q$ $$T_1 - T_2 = {(Q_1 - Q) - (Q_2 + Q)\over c_1m_1} = {Q_1 - Q_2 - 2Q\over c_1m_1} = {172431270 - 2Q\over 2090\times 916,7}$$ This happens every following moment. So, completing everything into one formula, $$Q = {58{172431270 - 2Q_{moved} \over 2090\times 916,7} \Delta t}$$

My question: Is there any way to do sum of all moments in 10 seconds to get heat transmitted ($Q$) between two solids? The only problem I have there is that $T_1$ - $T_2$ is not constant. I can't evaluate it without using $Q$.

Rewriting your starting equation slightly gives:

$$\frac{dQ}{dt} = A \Delta T$$

where I've merged all the constants into the the single constant $A$ so $A = \lambda S/L$. What we'd like is for the right hand side to be a function of $Q$, to make this a differential equation for $Q$. To do this note that if a quantity of heat $Q$ has flowed from hot to cold then the temperature of the hot side has decreased by $Q/cm$ and the temperature of the cold side has increased by $Q/cm$, so $\Delta T$ has decreased by $2Q/cm$. Putting this into our equation gives:

$$\frac{dQ}{dt} = A \left( \Delta T_0 - \frac{2Q}{cm} \right)$$

where $\Delta T_0$ is the initial temperature difference of 90°.

To make the algebra a bit easier note that if we wait for the temperatures to equalise, the quantity of heat that will have flowed is:

$$Q_\infty = \frac{\Delta T cm}{2}$$

and substituting for $\Delta T$ in our equation gives:

$$\frac{dQ}{dt} = A \frac{2}{cm} \left( Q_\infty - Q \right)$$

or:

$$\frac{dQ}{dt} = B \left( Q_\infty - Q \right)$$

where the new constant $B = 2\lambda S/cmL$. The solution to this is:

$$Q(t) = Q_\infty \left(1 - e^{-Bt} \right)$$

NB this assumes that the temperatures are uniform on the hot and cold sides, for example they could be rapidly stirred liquids. If the two sides are solid blocks then because heat flows at a finite speed within each block you get a complicated temperature profile. To calculate $Q$ in this case you'd need to solve the heat equation.