Complete Question
An unknown material, m1 = 0.41 kg, at a temperature of T1 = 86 degrees C is added to a Dewer (an insulated container) which contains m2 = 1.7 kg of water at T2 = 22 degrees C. Water has a specific heat of cw = 4186 J/(kg⋅K). After the system comes to equilibrium the final temperature is T = 30.3 degrees C.
Part (a) Input an expression for the specific heat of the unknown material.
Part (b) What is the specific heat in J/(kg⋅K)?
Answer:
a
[tex]c = \frac{m_2 * c_w (T - T_2)}{m_1 * (T_1 - T) }[/tex]
b
[tex]c = 2587.14 \ J /(kg \cdot K)[/tex]
Explanation:
From the question we are told that
The mass of the material is [tex]m_1 = 0.41 \ kg[/tex]
The temperature is [tex]T_1 = 86 ^oC[/tex]
The mass of water is [tex]m_2 = 1.7 \ kg[/tex]
The temperature of water is [tex]T_2 = 22^oC[/tex]
The specific heat of water is [tex]c_w = 4186 \ J/(kg\cdot K)[/tex]
The temperature of the system is [tex]T= 30 .3^o C[/tex]
Generally heat lost by the unknown material = heat gained by water
Generally the heat gained by water is mathematically represented as
[tex]Q_2= m_2 * c_w (T - T_2)[/tex]
=> [tex]Q_2= 1.7 * 4186 (30.3 - 22)[/tex]
=> [tex]Q_2= 59064.46 \ J[/tex]
Generally the heat lost by the unknown material is mathematically represented as
[tex]Q_1= m_1 * c* (T_1 - T)[/tex]
[tex]m_2 * c_w (T - T_2) = m_1 * c* (T_1 - T)[/tex]
=> [tex]c = \frac{m_2 * c_w (T - T_2)}{m_1 * (T_1 - T) }[/tex]
Here c is the specific heat capacity of the unknown material
=> [tex]Q_1= 0.41 * c (86 - 30.3)[/tex]
=> [tex]Q_1= 22.83 c[/tex]
So
[tex]59064.46 = 22.83 c[/tex]
=> [tex]c = \frac{59064.46 }{ 22.83}[/tex]
=> [tex]c = 2587.14 \ J /(kg \cdot K)[/tex]
