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if a 3.1g ring is heated using 10.0 calories, its temperatures rises 17.9C calculate the specific heat capacity of the ring

2 Answers

4 votes

Final answer:

The specific heat capacity of a 3.1g ring that is heated with 10.0 calories causing a temperature rise of 17.9°C is approximately 0.1801 cal/g°C.

Step-by-step explanation:

To calculate the specific heat capacity of the ring, we use the formula:

q = mcΔT

where:

  • q is the heat energy absorbed or released (in calories),
  • m is the mass of the substance (in grams),
  • c is the specific heat capacity (in cal/g°C), and
  • ΔT is the change in temperature (in °C).

Here, we are given that:

  • q = 10.0 calories,
  • m = 3.1 grams, and
  • ΔT = 17.9°C.

Now, we need to solve for c:

c = q / (mΔT)

Plugging in the known values:

c = 10.0 cal / (3.1 g × 17.9°C)

c = 10.0 cal / (55.49 g°C)

c ≈ 0.1801 cal/g°C

The specific heat capacity of the ring is approximately 0.1801 cal/g°C.

User Kristian Mo
by
8.8k points
3 votes
so what you do is you use the formula shown below:

specific heat capacity = energy required / (mass * change in temperature)

here ,
energy required = 10.0°C Note that cal. is short form for °C
mass (m) = 3.1g
change in temperature (ΔT) = 17.9°C Note that "ΔT" means change in temperature

So, plugging the values into the formula, we get,

Specific heat capacity=
(10)/(3.1*17.9)

=
0.1802126509

= 0.1802 cal./g°C i rounded the answer to the fourth decimal point




User Dean Michael
by
8.2k points

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