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.

1 answer

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:

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