A 50.3 g piece of aluminum (specific heat = 0.930 J/g･°C) will change its temperature from 23.0°C to 67.0°C while absorbing 2.06 × 10³ J of heat.
When a material absorbs heat, its temperature increases. We can calculate the amount of heat (Q) absorbed using the following expression.
[tex]Q = c \times m \times \Delta T[/tex]
where,
- ΔT: change in the temperature
[tex]Q = c \times m \times \Delta T\\Q = \frac{0.930J}{g. \° C } \times 50.3 g \times (67.0 \° C - 23.0 \° C) = 2.06 \times 10^{3} J[/tex]
A 50.3 g piece of aluminum (specific heat = 0.930 J/g･°C) will change its temperature from 23.0°C to 67.0°C while absorbing 2.06 × 10³ J of heat.
You can learn more about heat absorption here: https://brainly.com/question/15872403
