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If a fluid of mass 1 is heated for 2 minutes in a 2 kettle and its temperature increases by 50°C, what is the specific heat capacity of the fluid?

User Moini
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1 Answer

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Step-by-step explanation:

We can use the formula:

Q = mcΔT

where Q is the heat energy transferred, m is the mass of the fluid, c is the specific heat capacity, and ΔT is the change in temperature.

Since the fluid is heated in a kettle, we can assume that the heat energy transferred (Q) is equal to the heat energy absorbed by the fluid. Therefore:

Q = mcΔT

We are given that the mass of the fluid is 1 and the change in temperature is 50°C. We need to find the specific heat capacity (c).

Let's rearrange the formula to solve for c:

c = Q / (mΔT)

We need to find the value of Q. To do this, we can use the formula:

Q = Pt

where P is the power of the kettle and t is the time it is turned on. We are given that the kettle is turned on for 2 minutes. However, we need to find the power of the kettle.

We can assume that all the electrical energy supplied to the kettle is converted into heat energy, i.e.,:

electrical energy = heat energy

Therefore:

P × t = mcΔT

We can rearrange this formula to solve for P:

P = mcΔT / t

Now we can substitute the given values into the formula:

P = mcΔT / t = c × 1 × 50 / 2 = 25c

We are given that the fluid is heated for 2 minutes, so we can assume that the power of the kettle is constant during this time. Therefore:

electrical energy = Pt = 25ct

This electrical energy is converted into heat energy, i.e.,:

electrical energy = heat energy = mcΔT

Substituting the given values and solving for c:

c = Q / (mΔT) = (25ct) / (1 × 50) = 0.5t

Therefore, the specific heat capacity of the fluid is 0.5t.

User Ya Zhuang
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