## Cooling a CPU

A square silicon chip with a thermal conductivity of 150 W/m∙K is of width w = 7 mm on a side and of thickness t = 2 mm. The chip is mounted in a substrate such that its side and back surfaces are insulated, while the front surface is exposed to a coolant. If 60 W are being dissipated in circuits mounted to the back surface of the chip, what is the steady-state temperature difference between back and front surfaces?

##
__
__**Hint**

**Hint**

Fourier's Law of Conduction:

$$$P=\dot{Q}=kA\frac{\Delta T}{t}$$$

where
$$P$$
is power,
$$\dot{Q}$$
is the rate of heat transfer,
$$k$$
is the thermal conductivity,
$$A$$
is the surface area perpendicular to the heat transfer direction,
$$\Delta T$$
is the change in temperature, and
$$t$$
is the thickness.

##
__
__**Hint 2**

**Hint 2**

Because the silicon chip is a square profile:

$$$A=w^2$$$

where
$$w$$
is the width.

*Assumptions*: (1) Steady state conditions, (2) Constant properties, (3) Uniform heat dissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction chip.

All the electrical power dissipated at the chip's back surface is transferred by conduction. From Fourier's Law of Conduction:

$$$P=\dot{Q}=kA\frac{\Delta T}{t}$$$

where
$$P$$
is power,
$$\dot{Q}$$
is the rate of heat transfer,
$$k$$
is the thermal conductivity,
$$A$$
is the surface area perpendicular to the heat transfer direction,
$$\Delta T$$
is the change in temperature, and
$$t$$
is the thickness.

Rearranging the formula to solve for the temperature difference:

$$$\Delta T=\frac{t\cdot P}{kw^{2}}=\frac{0.002m\cdot 60W}{150W/m\cdot K(0.007m)^{2}}=16.33K$$$

16.33 K