Question

125.0 g of an unknown substance is heated to 97.0°C. It is then placed in a calorimeter than contains 250g of water with an initial temperature of 20.0°C. The final temperature reached by the calorimeter is 23.5°C. What is the specific heat of the unknown substance? The specific heat of water is 4.18 J/ (°C × g).

Answer 1

The answer should just be in your units. 250gX 3.5 C X 4.18 will give you the joules supplied by the substance. Divide that by 125.0g X 97-23.5 and you should have the specific heat. It has been a long times since ive done one of these.

Answer 2

The method above conceptually correct, but it has some errors..

you need to use the formula: \(\sf q=m*C_p*\Delta T\) twice.

where q is the heat absorbed/released
m is the mass of the substance in question (water or the unknown in this case.
\(C_p\) is the specific heat capacity
\(\Delta T\) is the change in temperature, this can be expanded to \(\Delta T=T_f-T_i\), where \(T_f\) is the final temp and \(T_i\) is the initial temp.

The thermal energy (heat) from the metal will be absorbed by the water (heat always flows spontaneously from high to low temps). The conventions are negative for release (exothermic processes) and positive for absorption (endothermic processes).

this means: \(\sf -q_{metal}=q_{water}\)

substituting the formula above, we get:

\(\sf m_{metal}*C^{metal}_p*(T_f-T^{metal}_i)=m_{water}*C^{water}_p*(T_f-T^{water}_i)\)

(note that the final temperature \(T_f\) is the same.

Solve for the specific heat capacity of the metal, \(\sf C^{metal}_p\):

\(\large \sf C^{metal}_p=\dfrac{m_{water}*C^{water}_p*(T_f-T^{water}_i)}{m_{metal}*(T_f-T^{metal}_i)}\)

plug in and solve (don't forget to check the units)

Answer 3

i forgot to include the negative sign,

it should look like this:

\(\sf \large \color{red}-[ m_{metal}*C^{metal}_p*(T_f-T^{metal}_i)]=m_{water}*C^{water}_p*(T_f-T^{water}_i)\)

then,

\(\large \sf C^{metal}_p=\color{red}-\dfrac{m_{water}*C^{water}_p*(T_f-T^{water}_i)}{m_{metal}*(T_f-T^{metal}_i)}\)