Question

A sample of brass is put into a calorimeter (see sketch at right) that contains of water. The brass sample starts off at and the temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at . Calculate the specific heat capacity of brass according to this experiment. Be sure your answer is rounded to significant digits.

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The specific heat is
`c_b = 0.402 J / g \cdot ^oC`

Step-by-step explanation:

From the question we are told that

The mass of the sample is
`m = 54.4 \ g`

The mass of the water is
`m_w = 150.0 \ g`

The initial temperature of the sample is
`T_i = 95.1 ^oC`

The initial temperature of the water is
`T_(w_i) = 15^oC`

The final temperature of the water is
`T = 17.6 ^oC`

Note the final temperature of water is equal to the final temperature of brass sample

The pressure is
`P =1 \ atm`

Generally for according to the law of energy conservation

The heat lost by sample = The heat gain by water

The heat lost by brass sample is mathematically evaluated as

`H_L = m * c_b * [T_i - T]`

Where
`c_b` is the specific neat of the brass sample

The heat gained by water is mathematically evaluated as

`H_g = m_w *c_w * [T_w - T ]`

where
`c_w` is the specific heat of water which has a constant value of

`c_w = 4.186 joule/gram`

So

`H_L = H_g \ \equiv m* c_b * [T_i -T] = m_w * c_w * [T - T_w]`

substituting values

`52.4 * c_b * [95.1 - 17.6] = 150 * 4.186 * [ 17.6 - 15.0]`

`c_b = 0.402 J / g \cdot ^oC`