Somebody please help me. A friend of mine is trying to figure this out and I don't know how to help her because I haven't learned this.
A student heats a 15.0 gram metallic sphere of unknown composition to a temperature of 98°C. The sphere is transferred to a calorimeter containing 100. mL of water at a temperature of 25.0°C. The student observes that the resulting temperture of both the water and the object is 27.1°C after the object is submerged.
1. Describe, in terms of the object and the water, the flow of heat energy that took place during the experiment.
2. Calculate the amount of heat energy gained by the water in the calorimeter.
3. Using the quantity of heat calculated in the previous question, determine the specific heat of the object.
2. Q = mcΔT?
I don't know how to calculate this, please help me!
With two anonymous names I don't know the difference; however, I assume the first one is answering the question 2 and the others one is saying s/he doesn't know how to use it.
q = heat gained
m = mass water. 100 mL water has a mass of 100 grams.
c = specific heat water which is 4.184 J/g*C
delta T is change in T which goes from 25.0 C to 27.1 C or 2.1.
887.8
1. To describe the flow of heat energy that took place during the experiment, we can apply the principle of conservation of energy. Heat energy flows from the hotter object to the colder object until both objects reach thermal equilibrium, where their temperatures are equal. In this case, the metallic sphere is initially heated to a temperature of 98°C and the water in the calorimeter is initially at a temperature of 25.0°C.
When the heated object is transferred to the water, heat energy transfers from the sphere to the water until both reach the final temperature of 27.1°C. The heat energy flows from the sphere, which has a higher initial temperature, to the water, which has a lower initial temperature. This transfer of heat causes a decrease in the temperature of the sphere and an increase in the temperature of the water. Once both reach 27.1°C, they are in thermal equilibrium, and no more heat transfer occurs.
2. To calculate the amount of heat energy gained by the water in the calorimeter, we can use the equation:
Q = m * c * ΔT
where Q is the heat energy gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given that the mass of water is 100.0 mL (which we can convert to grams since 1 mL of water weighs approximately 1 gram) and the change in temperature is (27.1 - 25.0) °C, we can substitute these values into the equation:
Q = (100.0 g) * (4.18 J/g°C) * (27.1 - 25.0) °C
Calculating this will give you the amount of heat energy gained by the water.
3. To determine the specific heat of the object, we can use the equation:
Q = m * c * ΔT
where Q is the heat energy gained by the water (which we calculated in the previous question), m is the mass of the object, c is the specific heat capacity of the object (what we want to determine), and ΔT is the change in temperature of the object.
Here, we know the mass of the object, which is given as 15.0 grams, and we know the change in temperature of the object, which is (27.1 - 98) °C. We also know the heat energy gained by the water from the previous question.
We can rearrange the equation to solve for the specific heat capacity (c):
c = Q / (m * ΔT)
Substituting the known values:
c = (heat energy gained by the water) / (15.0 g * (27.1 - 98) °C)
Calculating this will give us the specific heat capacity of the object.