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
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;
- Q = quantity of heat absorbed or released (J)
- m = mass of iron (g)
- c = specific heat capacity
- ∆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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