The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 47 calories of heat. What is the specific heat of iron?

Respuesta :

Q  =  mc(θ₂-θ₁)

47 calories =  10 g *c*(50.4 - 25)

47cal = 10*c* 25.4

47 /(10*25.4) = c

0.185 = c

Specific heat of iron = 0.185 cal/g°C 

The specific heat of iron with a mass of 10.0g that changed from 50.4°C to 25.0°C with the release of 47 calories is 0.19 J/cal°C.

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

  • The specific heat capacity of of substance can be calculated using the following expression:

Q = m × c × ∆T

c = Q ÷ (m∆T)

Where;

  1. Q = quantity of heat absorbed or released (J)
  2. m = mass of iron (g)
  3. c = specific heat capacity
  4. ∆T = change in temperature (°C)

  • According to this question, the temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 47 calories of heat.

c = 47 ÷ {10 × (50.4 - 25)}

c = 47 ÷ 254

c = 0.19 J/cal°C

  • Therefore, the specific heat of iron with a mass of 10.0 g that changed from 50.4°C to 25.0°C with the release of 47 calories is 0.19 J/cal°C.

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