Initially you have 204 g grams of solid thallium at is at its melting point of 304ºC .
How many grams of liquid thallium will be present after 476 joules of heat have been added.
specific heat liquid thallium is 0.13j/g ºC; heat fusion = 21 j/g; boiling point thallium is 1457ºC; heat vaporization = 795 j/g;
heat to convert solid Tl to liquid Tl at 304 C.
q1 = mass Tl x heat fusion = approx 4300 J; therefore, there is not enough heat to melt all of the Tl. How much can we melt? That's
476 J = mass Tl x 21 J/g
mass Tl = 476/21 = about 23 g
To solve this problem, we need to consider the different stages of thallium from solid to liquid, as well as the heat energy involved in each stage.
1. First, we need to calculate the amount of heat energy required to raise the temperature of the solid thallium from its melting point (304ºC) to its boiling point (1457ºC).
Q1 = mass * specific heat * change in temperature
Q1 = 204g * 0.13j/g ºC * (1457ºC - 304ºC)
2. Next, we need to calculate the amount of heat energy required for thallium to undergo the phase change from solid to liquid at its melting point.
Q2 = mass * heat fusion
Q2 = 204g * 21j/g
3. Now, we sum up the total heat energy required for these two steps to determine the heat energy required to entirely convert solid thallium into liquid thallium.
Total heat energy = Q1 + Q2
4. After calculating the total heat energy, we can determine the amount of liquid thallium produced based on the heat energy provided.
Q = mass * heat vaporization
476j = mass * 795j/g
5. Rearrange the equation to solve for mass:
mass = 476j / 795j/g
By substituting the value of mass into the equation, we can find the answer to the question.