In a carnot engine, the temperature of reservoir is $$527°$$C and that of sink is $$200$$ K. If the work done by the engine when it transfers heat from reservoir to sink is $$12000$$ kJ, the quantity of heat absorbed by the engine from reservoir is ______ $$\times 10^6$$ J.

We have a Carnot engine with reservoir temperature 527°C and sink temperature 200 K. The work done is 12000 kJ.

Converting the temperatures to Kelvin, we get $$T_1 = 527 + 273 = 800 \text{ K}$$ and $$T_2 = 200 \text{ K}.$$

The efficiency of the Carnot engine is given by $$\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{200}{800} = 1 - \frac{1}{4} = \frac{3}{4}.$$

Since efficiency is the ratio of work done to heat absorbed, $$\eta = \frac{W}{Q_1}.$$ Solving for the heat absorbed, $$Q_1 = \frac{W}{\eta} = \frac{12000}{\frac{3}{4}} = 12000 \times \frac{4}{3} = 16000 \text{ kJ}.$$

Converting to joules yields $$Q_1 = 16000 \text{ kJ} = 16 \times 10^6 \text{ J}.$$

The quantity of heat absorbed by the engine from the reservoir is 16 $$\times 10^6$$ J.