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
![c = (m_2 * c_w (T - T_2))/(m_1 * (T_1 - T) )](https://img.qammunity.org/2021/formulas/physics/high-school/6mttck778556fe9lxtu18sbdp08fugdmkn.png)
b
![c = 2587.14 \ J /(kg \cdot K)](https://img.qammunity.org/2021/formulas/physics/high-school/r9e5vr46x671bbo42rcxntvgmwuvjuh52o.png)
Step-by-step explanation:
From the question we are told that
The mass of the material is
![m_1 = 0.41 \ kg](https://img.qammunity.org/2021/formulas/physics/high-school/mb62fjaz0u06g4crpmn7rwlhj4np4aulgo.png)
The temperature is
![T_1 = 86 ^oC](https://img.qammunity.org/2021/formulas/physics/high-school/e18o964kxji0zd4fxa41xchim77cvtmb5q.png)
The mass of water is
![m_2 = 1.7 \ kg](https://img.qammunity.org/2021/formulas/physics/high-school/igryyvtilzk0omfxtiae7p5s2b9d7130j1.png)
The temperature of water is
![T_2 = 22^oC](https://img.qammunity.org/2021/formulas/physics/high-school/rgxm5k6habyy8wukcaubfjm7kjk66oh17e.png)
The specific heat of water is
![c_w = 4186 \ J/(kg\cdot K)](https://img.qammunity.org/2021/formulas/physics/high-school/yuuo3xmrszk39rqhl8lxfquwhuk3qetftp.png)
The temperature of the system is
![T= 30 .3^o C](https://img.qammunity.org/2021/formulas/physics/high-school/jdwbn20rz941z8ccc1hpmb2qljnlhu2glr.png)
Generally heat lost by the unknown material = heat gained by water
Generally the heat gained by water is mathematically represented as
![Q_2= m_2 * c_w (T - T_2)](https://img.qammunity.org/2021/formulas/physics/high-school/qksww57bmmndtu4cgqsvm7t8qcxr4j1cbi.png)
=>
![Q_2= 1.7 * 4186 (30.3 - 22)](https://img.qammunity.org/2021/formulas/physics/high-school/zahf262c8smy2zlasyx618iuu36jk9uunh.png)
=>
![Q_2= 59064.46 \ J](https://img.qammunity.org/2021/formulas/physics/high-school/6qonra2x64zjnsg6103pnmvec8o186kzeq.png)
Generally the heat lost by the unknown material is mathematically represented as
![Q_1= m_1 * c* (T_1 - T)](https://img.qammunity.org/2021/formulas/physics/high-school/ep6xk3mqg43j4f0r8vvnn6jbirtw1kzrp0.png)
![m_2 * c_w (T - T_2) = m_1 * c* (T_1 - T)](https://img.qammunity.org/2021/formulas/physics/high-school/r7594mq0rogwcksux9ohgxv6ngrb2rb06b.png)
=>
![c = (m_2 * c_w (T - T_2))/(m_1 * (T_1 - T) )](https://img.qammunity.org/2021/formulas/physics/high-school/6mttck778556fe9lxtu18sbdp08fugdmkn.png)
Here c is the specific heat capacity of the unknown material
=>
![Q_1= 0.41 * c (86 - 30.3)](https://img.qammunity.org/2021/formulas/physics/high-school/ie7jsbohlum77fnfi99wx20qlsbf1cguod.png)
=>
So
![59064.46 = 22.83 c](https://img.qammunity.org/2021/formulas/physics/high-school/wz0jy4kfbpbywylimn2znd71w2ekunz6qy.png)
=>
![c = (59064.46 )/( 22.83)](https://img.qammunity.org/2021/formulas/physics/high-school/hue9sxavjlas6fw43h094voidm2umvq4jy.png)
=>
![c = 2587.14 \ J /(kg \cdot K)](https://img.qammunity.org/2021/formulas/physics/high-school/r9e5vr46x671bbo42rcxntvgmwuvjuh52o.png)