A 100 g glass container is at 10.0 °C. 200 g of water at 90.0 °C is added to the glass container. What is the final temperature of the water and the glass, in °C? (specific heat of water = 1.00 cal/g °C, specific heat of glass = 0.200 cal/g °C)

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Quizplus · Accepted Answer

To find the final temperature, we use the principle of conservation of energy, where the heat lost by the water equals the heat gained by the glass container. Let $T_f$ be the final temperature.Heat lost by water = Heat gained by glass$$m_{water}c_{water}(T_{initial, water} - T_f) = m_{glass}c_{glass}(T_f - T_{initial, glass})$$Substituting the given values:$$200 \times 1.00 \times (90.0 - T_f) = 100 \times 0.200 \times (T_f - 10.0)$$$$200 \times (90.0 - T_f) = 20 \times (T_f - 10.0)$$$$18000 - 200T_f = 20T_f - 200$$$$220T_f = 18200$$$$T_f = \frac{18200}{220} = 82.7^\circ C$$Therefore, the final temperature of the water and the glass is 82.7°C.

A 100 G Glass Container Is at 10