A 110.0 g piece of molybdenum metal is heated to 100.0 °C and pla...

Question

A 110.0 g piece of molybdenum metal is heated to 100.0 °C and placed in a calorimeter that contains 150.0 g of water at 24.6 °C. The system reaches equilibirum at a final temeprature of 28.0 °C. Calcualte the specific heat of molybdenum metal in J/g·°C. The specific heat of water is 4.18 J/g·°C

Accepted Answer

Welcome back everyone in this example, we're told that in a calorie meter with 200 g of water at 25.7 degrees Celsius is 140 g of chromium metal placed at 73 degrees Celsius at a final temperature of 29 degrees Celsius, the system achieves equilibrium. If the specific heat of water is 4.18 joules per gram degree Celsius. What is the specific heat of chromium metal in jewels? Para grams, times degree Celsius. So in this prompt, we are relating the heat of one substance being are chromium metal to the heat of a second substance, which we will say is our water filling up our calorie meter. And whenever we have the heat of two substances related, one substance is going to be gaining heat while the other loses heat. So we're going to need a negative sign here. And what we should recognize from the prompt is that we have our chromium metal at a higher temperature where once the system achieves equilibrium, the chromium metal is going to be also that temperature of the system being 29 degrees Celsius. So it appears that the chromium metal should be losing heat where our water begins at a lower temperature than at equilibrium being 25.7 degrees Celsius. So the water is going to be gaining heat when it ends at 29 degrees Celsius at equilibrium. So we're going to have, since water is gaining heat, it's going to have a negative sign here, since the heat of our chromium metal is being related to the heat of water. Now you can really use the negative sign on any side of the equation. But we're going to follow it in this order for this case. So for the specific use of chromium, we're going to plug in our terms for specifically capacity. So we're relating the mass of our chromium metal given in the prompt as 140g multiplied by the specific heat of chromium metal which was solving force who are plugging in as the variable C. And then multiplied by the difference in temperature delta T. So we're going to recall that from the prompt. The final temperature is given as the equilibrium temperature of 29 degrees Celsius subtracted from its initial temperature given in the prompt as 73 degrees Celsius. And so now relating this to the specific heat Formula for water, we have the massive water given in the pumped as 200 g multiplied by the specific heat of water given in the prompt as 4.18 joules per grams times degree Celsius and then multiplied by the difference in temperature where we have a final temperature for the water being the equilibrium temperature of the system being 29 degrees Celsius. Subtracted from the initial temperature given in the prompt for water as 25.7 degrees Celsius. And so simplifying our right hand side first, we're going to take the product and the difference there. And what we're going to have is that our left hand side. 140 g times the specific key of chromium times the difference in temperature is equal to our right hand side, which simplifies to negative 2000 and sorry about not keeping that negative sign that we still want that negative sign there. So it's negative 2758.8. And as far as units will be able to cancel out grams as well as degrees Celsius, leaving us with jewels as our final unit on the right hand side. So we have 2000 negative 2758.8 jewels. And now simplifying further, we're going to divide both sides by our product of g times 29 degrees Celsius minus 73 degrees Celsius so that those terms cancel out on the right hand side And then on the left hand side. So on the right hand side we're dividing by again, 140g times 29.0°C -73.0°C. And so now we're just going to have our specific heat of chromium fully isolated so that we set it equal to the product of our quotient here, which is going to simplify in our calculators to a value of .4478 tools programs, times degrees Celsius. And since in our prompt, our minimum number of sig figs was. Three sig figs. We're going to simplify this to around this 23 sig figs as 230.448 jewels. Programs, times degrees Celsius. So this would be actually our final answer for our specific heat capacity of our chromium metal. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

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