Specific Heat Capacity, Heat Energy, Thermodynamics Physics

[RCF]

Homework Statement

A block of ice has a mass of 150g and a temperature of -4°. The ice is melted by supplying 60KJ heat energy. Determine the final temperature of the melted water.

Homework Equations

Equations i think can be used:

Q = ml and H = mcΔθ

The Attempt at a Solution

I have attempted the solution with even different specific heat capacities but i am getting very high temperature which does not make a lot of sense. Can you please help me out.

 

[epenguin]

Have you heard of latent heat? (if they still call it that. Anyway heat of melting.)

 

[RCF]

yes but how to solve it using latent heat studies?

 

[epenguin]

Well if you know what latent heat is calculate the amount of heat energy that melts that amount of ice, which is without change of temperature, then the rest of the supplied 60 kJ goes into heating the water and ice, then use specific heats to calculate how much those kJ raise the temperature. You have presumably been supplied with the relevant physical parameters to use, else you will have to find them out.

 

[Nickie]

As a scientist, it is important to understand and apply the principles of thermodynamics in order to accurately solve problems involving heat energy and temperature changes. In this case, we are given a block of ice with a mass of 150g and a temperature of -4°, and we are told that it is melted by supplying 60KJ of heat energy. Our task is to determine the final temperature of the melted water.

To solve this problem, we can use the equation Q = ml, where Q is the heat energy, m is the mass, and l is the specific heat capacity. We can also use the equation H = mcΔθ, where H is the heat energy, m is the mass, c is the specific heat capacity, and Δθ is the change in temperature.

Since we are given the mass and heat energy, we can rearrange the first equation to solve for the specific heat capacity, l.

l = Q/m

Plugging in the values, we get:

l = (60,000J)/(0.150kg)

l = 400,000 J/kg

Now, using the second equation, we can solve for the final temperature, Δθ.

Δθ = H/(mc)

Plugging in the values, we get:

Δθ = (60,000J)/((0.150kg)(400,000 J/kg))

Δθ = 0.25 K

Therefore, the final temperature of the melted water is -4° + 0.25 K = -3.75°.

It is important to note that the specific heat capacity of water is not a constant value and can vary depending on the temperature. However, for simplicity, we have assumed a constant specific heat capacity in this problem. Additionally, our final temperature of -3.75° may seem high, but it is important to remember that we are dealing with a small amount of ice (150g) and a large amount of heat energy (60KJ), so the temperature change will be significant.

In conclusion, by using the principles of thermodynamics and the equations Q = ml and H = mcΔθ, we can accurately determine the final temperature of the melted water in this problem.