A refrigerator consumes an average 35 W power to operate between temperature -10°C to 25°C. If there is no loss of energy then how much average heat per second does it transfer?

We begin by converting the given Celsius temperatures into Kelvin, because the thermodynamic formulae for a Carnot refrigerator demand absolute temperatures.

We have
$$T_{\text{cold}} = -10^{\circ}{\rm C} = -10 + 273 = 263\ {\rm K}$$
and
$$T_{\text{hot}} = 25^{\circ}{\rm C} = 25 + 273 = 298\ {\rm K}.$$

For an ideal, perfectly reversible refrigerator the coefficient of performance (COP) is given by the Carnot expression

$$\text{COP} = \frac{Q_{\text{cold}}}{W} = \frac{T_{\text{cold}}}{T_{\text{hot}} - T_{\text{cold}}},$$

where $$Q_{\text{cold}}$$ is the heat absorbed per second from the cold chamber and $$W$$ is the mechanical work (power) supplied per second.

Substituting the numerical values, we obtain

$$\text{COP} = \frac{263}{298 - 263} = \frac{263}{35}.$$

Carrying out the division gives

$$\text{COP} = 7.514.$$

Now, the power input is given as $$W = 35\ {\rm W} = 35\ {\rm J\,s^{-1}}.$$
Using the relation $$Q_{\text{cold}} = \text{COP} \times W$$ we can find the average heat drawn from the cold compartment each second.

So

$$Q_{\text{cold}} = 7.514 \times 35\ {\rm J\,s^{-1}}.$$

Multiplying, we get

$$Q_{\text{cold}} = 263\ {\rm J\,s^{-1}}\;(\text{approximately}).$$

This quantity represents the average heat that the refrigerator transfers per second from the low-temperature space when there are no energy losses.

Hence, the correct answer is Option C.