A 25.0 g piece of granite at 100.0°C was added to 100.0 g of wate...

Question

A 25.0 g piece of granite at 100.0°C was added to 100.0 g of water of 25.0°C, and the temperature rose to 28.4°C. What is the specific heat capacity of the granite? (The specific heat capacity for water is 4.18 J/(g•°C).) (LO 9.10)
(a) 0.563 J/(g•°C)
(b) 1.53 J/(g•°C)
(c) 0.992 J/(g•°C)
(d) 0.794 J/(g•°C)

Accepted Answer

Hello everyone today. We are being given the falling problem a 35 g sample of an unknown material at 90 degrees Celsius is submerged in 100 g of water at 30 degrees Celsius, causing the temperature of water to raise to 42.5 degrees Celsius determine the specific heat capacity of the material. So we want to know what we're going to be working with here. And according to our principles, we are going to be working with a concept known as thermal equilibrium. And so what is this state? This says that the material that we are working with and the water. So the material and the water will have the same final temperature. So they will have the same final temperature. And so we have to note this actual temperature of the water. So we have a final temperature of our water or H. 20. So you have T. F. Of our H 20. That is going to equal the final temperature of our material. Which as noted in the question was 42.5°C. And so we'll save that number for later. We also need to note that the heat of our material is going to be equal to the opposite for our water. So the heat for our water will be the same as the material. Except it will be the opposite. So in this case the material heat is positive. And so our heat for our our energy for our water will be negative. And so next, what does our Q. Stand for that is our heat more specifically, that is our specific heat capacity formula. That says that M. R. Mass times our specific heat times our temperature. And so we have that for our material. And we're going to equal that to our negative specific heat capacity. Using the same variables. We're going to plug in our units and values and then we're gonna ultimately solve for our specific heat capacity. So our mass for our material is 35 g per the question. Our specific heat capacity we do not have. So we'll just leave that as AC. And our change in temperature. It was noted that we had a final temperature of 42.5°C. But we started with the material at 90°C. We are then going to equal that to our negative mass of our water which was said to be 100 g per the question times the specific heat capacity of water which will be in your textbooks and it is 4.18 jewels per grams times Celsius. We're then gonna lastly multiply that By our change in our temperature for water. So the final temperature of our water was 42.5°C. And it started at 30°C. And so when we ultimately solve for our specific heat capacity we will get that C. Is equal to 3. jewels per grams times Celsius as our final answer. Ultimately I hope that this helped. And until next time

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