Question

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

Answer 1

• 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

Answer 2

About 3.5 kJ.

Step-by-step explanation:

Recall the heat transfer equation:

`\displaystyle q = mC\Delta T`

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:

`\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* 10^3 \text{ J} = 3.5\text{ kJ}\end{aligned}`

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