Question

8. An electrical resistor immersed in a liquid produces 65.0 W of electrical energy for 120
specific heat capacity of the liquid in this temperature range. Assume that negligible h
O
O
O
2.51 x 10³ J/(kg. C°)
1.67 x 106 J/(kg C°)
2.02 x 104 J/(kg. C°)

Answer 1

The specific heat capacity of the liquid is 2.51 x 10³ J/kg⁰C.

Specific heat capacity of the liquid

The specific heat capacity of the liquid is calculated as follows;

Q = mcΔФ

where;

• Q is quantity of heat
• c is specific heat capacity
• m is mass of the liquid
• ΔФ is change in temperature = 22.54 - 18.55 = 3.99 ⁰C

c = Q/mΔФ

c = (65 x 120)/(0.78 x 3.99)

c = 2.51 x 10³ J/kg⁰C

The complete question is below:

An electrical resistor immersed in a liquid produces 65.0 W of electrical energy for 120 s. The mass of the liquid is 0.780 kg and its temperature increases from 18.55°C to 22.54°C. a) Find the average specific heat capacity of the liquid in this temperature range.

Learn more about specific heat capacity here: https://brainly.com/question/16559442

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