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A 1 kg block of iron sits outside in the direct sun from 8am to 10am. At 8am the temperature of the block was 20.0°C and at 10am the temperature of the block was 27.5°C.
How much heat did the block of Iron absorb from the sun?
use q=mcΔT

Respuesta :

xKelvin

Answer:

About 3.5 kJ.

Explanation:

Recall the heat transfer equation:

[tex]\displaystyle q = mC\Delta T[/tex]

Where q is the heat; m is the mass of the substance; C is its specific heat; and ΔT is the change in temperature.

The given specific heat of iron is 462 J/kg -°C.

Substitute and evaluate. Ensure correct units:
[tex]\displaystyle \begin{aligned} q & = (1\text{ kg})\left(\frac{462\text{ J}}{\text{kg -$^\circ$C}}\right)(27.5\text{ $^\circ$C} - 20.0\text{ $^\circ$C}) \\ \\ & = (1\text{ kg})\left(\frac{462\text{ J}}{\text{kg -$^\circ$C}}\right)(7.5\text{ $^\circ$C}) \\ \\ & = 3.5\times 10^3 \text{ J} = 3.5\text{ kJ}\end{aligned}[/tex]

Therefore, the block of iron absorbed about 3.5 kJ of heat.

  • specific heat capacity of iron=462J/kg°C
  • Change in temperature=∆T=(27.5-20)=7.5°C

So

  • Q=mc∆T
  • Q=1(462)(7.5)
  • Q=3500J

The block absorbed 3500J of heat

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