A copper ball of mass 100 g is at a temperature $$T$$. It is dropped in a copper calorimeter of mass 100 g, filled with 170 g of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. $$T$$ is given by:
(Given: room temperature = 30°C, specific heat of copper = 0.1 cal g$$^{-1}$$ °C$$^{-1}$$)

We first recall the basic principle of calorimetry: in an isolated system, the total heat lost by the hotter objects equals the total heat gained by the colder ones, that is

$$\text{Heat lost} \;=\; \text{Heat gained}.$$

The data given are

$$\begin{aligned} \text{Mass of hot copper ball} & = 100 \,\text{g},\\[4pt] \text{Specific heat of copper} & = 0.1 \,\text{cal g}^{-1}{}^{\circ}\!{\rm C}^{-1},\\[4pt] \text{Initial temperature of water and calorimeter} & = 30^{\circ}\!{\rm C},\\[4pt] \text{Mass of copper calorimeter} & = 100 \,\text{g},\\[4pt] \text{Mass of water} & = 170 \,\text{g},\\[4pt] \text{Equilibrium temperature} & = 75^{\circ}\!{\rm C}. \end{aligned}$$

We take the specific heat of water to be the standard value

$$s_{\text{water}} = 1 \,\text{cal g}^{-1}{}^{\circ}\!{\rm C}^{-1}.$$

Heat lost by the hot copper ball as it cools from the unknown temperature $$T$$ to $$75^{\circ}\!{\rm C}$$:

$$ Q_{\text{lost}} = m_{\text{ball}}\;s_{\text{Cu}}\;(T - 75) = 100 \times 0.1 \times (T - 75) = 10\,(T - 75)\ \text{cal}. $$

Heat gained by the copper calorimeter as it warms from $$30^{\circ}\!{\rm C}$$ to $$75^{\circ}\!{\rm C}$$:

$$ Q_{\text{cal}} = m_{\text{cal}}\;s_{\text{Cu}}\;(75 - 30) = 100 \times 0.1 \times 45 = 450\ \text{cal}. $$

Heat gained by the water over the same temperature rise:

$$ Q_{\text{water}} = m_{\text{water}}\;s_{\text{water}}\;(75 - 30) = 170 \times 1 \times 45 = 7650\ \text{cal}. $$

Total heat gained by the colder parts of the system is therefore

$$ Q_{\text{gained}} = Q_{\text{cal}} + Q_{\text{water}} = 450 + 7650 = 8100\ \text{cal}. $$

Applying the calorimetry principle, we equate the two heats:

$$ Q_{\text{lost}} = Q_{\text{gained}} \;\;\Longrightarrow\;\; 10\,(T - 75) = 8100. $$

Solving this simple linear equation step by step, we divide by 10:

$$ T - 75 = \frac{8100}{10} = 810. $$

Adding $$75^{\circ}\!{\rm C}$$ to both sides gives the initial temperature of the copper ball:

$$ T = 810 + 75 = 885^{\circ}\!{\rm C}. $$

Thus the copper ball was originally at a temperature of $$885^{\circ}\!{\rm C}$$.

Hence, the correct answer is Option C.