In this example, we find the final temperature after a hot metal is dropped into cold water and they reach thermal equilibrium. Our problem reads, "A 1.75 kilogram lead block initially at 98.0 degrees Celsius is dropped into 435 grams of water initially at 21.4 degrees Celsius. What is the final temperature of the water and lead in degrees Celsius? The specific heat capacity of water and lead are 4.18 and 0.128 joules per gram degree Celsius, respectively. To solve this problem, we need to realize that the thermal energy absorbed by the water is released by the metal, so Q water equals -Q metal. This is the mathematical equivalent of saying that the energy lost by the metal is gained by the water. Our equation for Q is heat equals mass times specific heat capacity times change in temperature. We can plug in our values for water: we have 435 g of water, 4.18 joules per gram degree Celsius for the specific heat capacity, and the change in temperature is final temperature minus initial temperature, which is T final minus 21.4 degrees Celsius. For our metal, we want to make sure we convert 1.75 kilograms to grams, so we enter 1750 grams for the mass, 0.128 joules per gram degree Celsius for the specific heat capacity, and T final minus 98.0 degrees Celsius for the change in temperature. Next, we'll need to plug these expressions for Q water and Q metal into Q water equals -Q metal, and then we'll solve for T final. In order to solve for the final temperature, we can first use the distributive property on each side. We're multiplying what's outside the parentheses by T final and by the initial temperature. On the left, 435 grams times 4.18 joules per gram degree Celsius times T final is 1818.3 joules per degree Celsius times T final. 435 grams times 4.18 joules per gram degree Celsius times -21.4 degrees Celsius is 38,912 joules. On the right, -1750 grams times 0.128 joules per gram degree Celsius times T final is -224 joules per degree Celsius times T final. And -1750 grams times 0.128 joules per gram degree Celsius times -98 degrees Celsius is 21,952 joules. Now we can move all of the T finals to the left side and all of the numbers to the right side by adding 224 joules per degree Celsius times T final to both sides and adding 38,912 joules to both sides. This will bring all of the joules to the right side of the equation and the T finals to the left side of the equation. Combining coefficients, 1818.3 plus 224 is 2042.3 times T final. On the right, 21,952 plus 38,912 is 60,864 joules. Finally, we can divide both sides by 2042.3 joules per degree Celsius to find the final temperature, which is 29.8 degrees Celsius.