Question

An unknown material, m1 = 0.49 kg, at a temperature of T1 = 92 degrees C is added to a Dewer (an insulated container) which contains m2 = 1.1 kg of water at T2 = 21 degrees C. Water has a specific heat of cw = 4186 J/(kg⋅K). After the system comes to equilibrium the final temperature is T = 31 degrees C. Detemine the specific heat of unknown material.

Answer 1

Answer:

$c_u=1540.5J/kg^{\circ}K$

Explanation:

We know that heat relates to mass, specific heat and variation of temperature experimented because of this heat through the equation $Q=mc\Delta T=mc(T_f-T_i)$. The heat released by the unknown material is absorbed by water, so we have $Q_u=-Q_w$, and we can write:

$m_uc_u(T_{uf}-T_{ui})=-m_wc_w(T_{wf}-T_{wi})$

Since thermal equilibrium is reached we know that $T_{cf}=T_{wf}=T_f=31^{\circ}C=304^{\circ}K$, where we have added $273^{\circ}$ to convert the temperature from Celsius to Kelvin, as we must do. Since we want the specific heat of the unknown material, we do:

$c_u=-\frac{m_wc_w(T_f-T_{wi})}{m_u(T_f-T_{ui})}$

Which for our values is:

$c_u=-\frac{(1.1kg)(4186J/kg^{\circ}K)((304^{\circ}K)-(294^{\circ}K))}{(0.49kg)((304^{\circ}K)-(365^{\circ}K))}=1540.5J/kg^{\circ}K$