Determine surface area of heat exchange tube A heat exchanger is used to cool down a stream of $\pu{20 kg s-1}$ of water from $\pu{80 °C}$ to $\pu{60 °C}$. The cooling water enters at a rate of $\pu{12 kg s-1}$ and $\ce{20 °C}$. The overall heat transfer coefficient is $\pu{2 kW m-2 K-1}$ and the heat capacity of water is $\pu{4.183 kJ kg-1 K-1}$

Calculate the surface area.

I am stuck on what the last step of the calculations should be.
I began with calculating the outlet temperature for the cooling stream using that $q$ is constant, I set IN = OUT:
$$m_hC_pΔT_\mathrm{in} = m_cC_pΔT_\mathrm{out}\label{eq:1}\tag{1}$$
From $\eqref{eq:1}$ I got that $T_\mathrm{2out} = \pu{53 °C}$
Then I calculated the logarithmic temperature difference, which I got to be
$ΔT_\mathrm{ln} = \pu{-33.075 °C}$.
After this I am confused as to how I can get the surface area of the tube.
Edit
I did a heat balance, IN - OUT = 0:
$$(m_hC_pΔT_\mathrm{in} + m_cC_pΔT_\mathrm{in}) - (m_hC_pΔT_\mathrm{out} + m_cC_pΔT_\mathrm{out}) = 0$$
simplified into:
$$m_hC_pΔT_\mathrm{in} - m_hC_pΔT_\mathrm{out} - UAΔT_\mathrm{ln} = 0$$
However when I try to solve for A I get a negative value, which isn't correct.
The outlet temperature  should be 53.3.  The log-mean temperaturr difference should be positive.  Calculate the heat load and divide by the log-mean temperature difference and by the heat transfer coefficient to get the area.

ADDENDUM

This addendum is for the benefit of @MaxW (the downvoter).


The heat load $\dot{Q}$ of the heat exchanger is given by:  $$\dot{Q}=\dot{m}_HC(T_{H,in}-T_{H,out})=20(4.183)(80-60)=1673\ kW$$This heat load represents the rate of heat transfer from the hot fluid to the cold fluid.
The exit temperature of the cold stream is determined from $$\dot{m}_CC(T_{C,out}-T_{C,in})=12(4.183)(T_{C,out}-20)=\dot{Q}=1673$$


Solving for the exit temperature of the cold stream gives  $$T_{C,out}=53.33\ C$$
3.  The log-mean temperature difference is given by:
$$(\Delta T)_{LM}=\frac{(T_{H,in}-T_{C,out})-(T_{H,out}-T_{C,in})}{\ln{[(T_{H,in}-T_{C,out})/(T_{H,out}-T_{C,in})]}}=\frac{26.67-40}{\ln{[26.67/40]}}=32.88\ C$$


The heat transfer area A is determined from the equation:  $$U_oA(\Delta T)_{LM}=2.0A(32.88)=\dot{Q}=1673$$Solving this equation for the heat transfer area A gives:
$$A=25.4\ m^2$$


Questions?