Question

2kg of water of 40°C is mixed with 1kg of water at 80°C. Calculate the temperature of the mixure​

Answer 1

So, the equilibrium temperature after the first object mixed with second object is approximately 53.3°C.

Introduction

Hi! I will help you to solve Black's Principle. Black's Principle state that the heat absorbed by one object is always similar to the heat released by another. So, these objects will transfer heat into each other until they reach an equilibrium temperature. Usually, Black's principle used to find the equilibrium temperature after the interaction of two objects with the same masses but different types, with the same type but different masses, or different types and masses. The type of objects will affect the heat absorbed or released because each type has its specific heat characteristics (c).

Formula Used

So, mathematically Black's Principle can be expressed in the following equation:

`\sf{\bold{Q_1 = Q_2}}`

`\boxed{\sf{\bold{m_1 \cdot c_1 \cdot (T_1-T') = m_2 \cdot c_2 \cdot (T'-T_2)}}}`

With the following condition:

• 
  `\sf{m_1}` = mass of the first object
• 
  `\sf{m_2}` = mass of the second object
• 
  `\sf{c_1}` = specific heat of the first object
• 
  `\sf{c_2}` = specific heat of the second object
• 
  `\sf{T'}` = change in temperature value
• 
  `\sf{T_1}` = temperature of the first object
• 
  `\sf{T_2}` = temperature of the second object
• The first object has higher temperature than the second object.

Problem Solving

We know that:

• 
  `\sf{m_1}` = mass of the first object = 1 kg.
• 
  `\sf{m_2}` = mass of the second object = 2 kg.
• 
  `\sf{c_1 = c_2 = c}` = specific heat of water >> the value can ignored because its will devided by each other.
• 
  `\sf{T_1}` = temperature of the first object =
  `\sf{80^\circ}` C.
• 
  `\sf{T_2}` = temperature of the second object =
  `\sf{80^\circ}` C

What was asked ?

• 
  `\sf{T'}` = change in temperature value = ...°C

Step by step :

`\sf{m_1 \cdot c_1 \cdot (T_1-T')= m_2 \cdot c_2 \cdot (T'-T_2)}`

`\sf{1 \cdot \cancel c \cdot (80-T')= 2 \cdot \cancel c \cdot (T'-40)}`

`\sf{80 - T'= 2T' - 80}`

`\sf{-3T'= -160}`

`\sf{T'= (-160)/(-3)}`

`\sf{\bold{T' \approx 53.3^\circ C}}`

Conclusion

So, the equilibrium temperature after the first object mixed with second object is approximately 53.3°C.