Question

A 15 kg block of brass is heated from 295 K to 350. K. How much heat was absorbed by the brass? The specific heat of brass is 376 J/kg•K.

Answer 1

To calculate the heat absorbed by the brass block, you can use the formula:

\[
Q = mc\Delta T
\]

where:
- \( Q \) is the heat absorbed (in joules),
- \( m \) is the mass of the brass block (in kg),
- \( c \) is the specific heat capacity (in J/kg·K),
- \( \Delta T \) is the change in temperature (in K).

Given:
- \( m = 15 \) kg
- \( c = 376 \) J/kg·K
- Initial temperature \( T_i = 295 \) K
- Final temperature \( T_f = 350 \) K

First, calculate the change in temperature (\( \Delta T \)):

\[
\Delta T = T_f - T_i = 350 \, \text{K} - 295 \, \text{K} = 55 \, \text{K}
\]

Now, plug in the values to find \( Q \):

\[
Q = mc\Delta T = (15 \, \text{kg}) \times (376 \, \text{J/kg·K}) \times (55 \, \text{K})
\]

Calculating this gives:

\[
Q = 15 \times 376 \times 55
\]

First calculate \( 15 \times 376 \):

\[
15 \times 376 = 5640 \, \text{J/K}
\]

Now, multiply that by \( 55 \):

\[
Q = 5640 \, \text{J/K} \times 55 \, \text{K} = 310200 \, \text{J}
\]

Therefore, the total amount of heat absorbed by the brass is:

\[
\boxed{310200 \, \text{J}}
\]