Question

A student placed 25.421 g of metal in A coffee cup with exactly 50 g of water initial temperature of the metal was 99.5°C and the initial temperature of the water in the cup was 22.5°C if the final temp of the system with 31.8°C what is the specific heat of the metal

B) if the actual specific heat for the metal in question one was 1.35J/g c calculate the percentage error

Answer 1

Answer on the question #60227, Chemistry / General Chemistry

Question:

A student placed 25.421 g of metal in A coffee cup with exactly 50 g of water initial temperature of the metal was 99.5°C and the initial temperature of the water in the cup was 22.5°C if the final temp of the system with 31.8°C what is the specific heat of the metal

B) if the actual specific heat for the metal in question one was 1.35 J/g C calculate the percentage error

Solution:

A) If we assume no energy losses, then the quantity of heat that water received and the quantity of heat that metal provided are the same:

$$Q_{\text{water}} = -Q_{\text{metal}}$$

The heat is linked to the change of the temperature of the body in the following way:

$$Q = c m (T_2 - T_1),$$

where $c$ is the heat capacity, $m$ is the mass and $T_2$ and $T_1$ are the final and the initial temperature, respectively.

So, for the former equation we can write:

$$c_{\text{water}} m_{\text{water}} (T_2 - T_{1,\text{water}}) = -c_{\text{metal}} m_{\text{metal}} (T_2 - T_{1,\text{metal}})$$

The heat capacity of water is 1 cal/g °C, so we can rearrange the last equation:

$$c_{\text{metal}} = \frac{c_{\text{water}} m_{\text{water}} (T_2 - T_{1,\text{water}})}{m_{\text{metal}} (-T_2 + T_{1,\text{metal}})}$$

And finally,

$$c_{\text{metal}} = 0.27 \frac{\text{cal}}{\text{g}^{\circ}\text{C}}$$

B) To calculate this, we should convert the value in cal to J. As 1 cal is 4.184 J, it is very easy to see that 0.27 cal/g °C = 0.27·4.184 = 1.13 J/g °C

Then, the percentage error is:

$$x = \frac{1.35 - 1.13}{1.35} \cdot 100\% = 16\%$$

Answer: A) 0.27 cal/g °C, or 1.13 J/g °C B) 16%