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Irish coffeeYou make a cup of Irish coffee out of 300 g of coffee at 100 °C and 30 g of pure ethyl alcohol at 20 °C. One Joule is enough energy to produce a change of 1 °C in 0.42 g of ethyl alcohol (i.e., alcohol is easier to heat than water). What temperature is the final mixture? Adding up all the energy after mixing has to give the same result as the total before mixing. We let the subscript i stand for the initial situation, before mixing, and f for the final situation, and use subscripts c for the coffee and a for the alcohol. In this notation, we have total initial energy = total final energy E_{ci} + E_{ai} = E_{cf} + E_{af }. We assume coffee has the same heatcarrying properties as water. Our information about the heatcarrying properties of the two substances is stated in terms of the change in energy required for a certain change in temperature, so we rearrange the equation to express everything in terms of energy differences: E_{af}  E_{ai} = E_{ci}  E_{cf} . Using the given ratios of temperature change to energy change, we have E_{ci}  E_{cf} = (T_{ci}  T_{cf} )(m_{c})/(0.24 g) E_{af}  E_{ai} = (T_{af}  T_{ai} )(m_{a})/(0.42 g) Setting these two quantities to be equal, we have (T_{af}  T_{ai} )(m_{a})/(0.42 g) = (T_{ci}  T_{cf} )(m_{c})/(0.24 g) . In the final mixture the two substances must be at the same temperature, so we can use a single symbol T_{f} = T_{cf} = T_{af} for the two quantities previously represented by two different symbols, (T_{f}  T_{ai} )(m_{a})/(0.42 g) = (T_{ci}  T_{f} )(m_{c})/(0.24 g) . Solving for T_{f} gives


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