application of heat capacity

Question 1

Question:
A 200 g block of aluminum at an initial temperature of 20°C is heated until its temperature reaches 80°C. If the specific heat capacity of aluminum is
0.90 J/(g \ °C) 0.90J /(g\°C), how much heat energy was absorbed by the aluminum block during heating?

Answer 1

Solution:
To find the heat energy absorbed (( Q )), we can use the formula:
[ Q = m CDelta T ]
where:
- ( m ) is the mass of the substance (in grams),
- C_is the specific heat capacity of the substance (in °C)}
- ( \Delta T ) is the change in temperature (in °C).

Given:
- m = 200
C = 0.90
Delta T = 80°C - 20°C = 60°C

Substituting the given values into the formula:
[ Q = 200 x 0.90 x 60°C

Q = 200 \ x 0.90 x 60

Q = 10,800

Answer:
The aluminum block absorbed 10,800 ,of heat energy during heating.