
A new temperature scale uses X as a unit of temperature, where the numerical value of the temperature ${t_x}$ in this scale is related to the absolute temperature $T$ by ${t_x} = 3T + 300$ . If the specific heat of a material using this unit is $1400{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{X}}^{ - 1}}$ , find its specific heat in the S.I. system of units.
A) $4200{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$
B) $1400{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$
C) ${\text{466}}{\text{.7Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$
D) Impossible to determine from the given information.
Answer
582.6k+ views
Hint:The given relation between the temperature in the X system of units and the temperature in the S.I. system of units suggests that a change in temperature of 1 K in the S.I. system of unit will be reflected as a change in temperature of 3 K in the new system.
Complete step by step answer.
Step 1: Using the relation ${t_x} = 3T + 300$ find the value of ${t_x}$ when the temperature in S.I. unit changes from 0 K to 1 K and obtain the factor by which the temperature in the X system increases.
Given the relation for the temperature ${t_x}$ in the X scale as ${t_x} = 3T + 300$ .
For $T = 0{\text{K}}$ , ${t_x} = 3 \times 0 + 300 = 300{\text{X}}$
For $T = 1{\text{K}}$ , ${t_x} = 3 \times 1 + 300 = 303{\text{X}}$
Hence we change $\delta \left( {{t_x}} \right) = 303 - 300 = 3$ .
This implies that when there is a change of 1 K in the S.I. system, the corresponding change in the X system will be 3 K.
Step 2: Using the obtained factor of change of temperature, find the specific heat of the material in the S.I. system.
The specific heat of the material in the X system is given to be $1400{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{X}}^{ - 1}}$ .
The factor by which the temperature changes in the X system is obtained as $\delta \left( {{t_x}} \right) = 3$
So, the specific heat of the material in the S.I. unit can be obtained by $1400 \times 3 = 4200{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$
Hence the correct option is A.
Note:Specific heat of a material refers to the amount of heat required to raise the temperature of a material of mass of $1{\text{kg}}$ by $1{\text{K}}$. Since $\delta \left( {{t_x}} \right) = 3$, in the new system, the specific heat of the material will be the amount of heat required to change the temperature of the material of mass of $1{\text{kg}}$ by ${\text{3K}}$.
Complete step by step answer.
Step 1: Using the relation ${t_x} = 3T + 300$ find the value of ${t_x}$ when the temperature in S.I. unit changes from 0 K to 1 K and obtain the factor by which the temperature in the X system increases.
Given the relation for the temperature ${t_x}$ in the X scale as ${t_x} = 3T + 300$ .
For $T = 0{\text{K}}$ , ${t_x} = 3 \times 0 + 300 = 300{\text{X}}$
For $T = 1{\text{K}}$ , ${t_x} = 3 \times 1 + 300 = 303{\text{X}}$
Hence we change $\delta \left( {{t_x}} \right) = 303 - 300 = 3$ .
This implies that when there is a change of 1 K in the S.I. system, the corresponding change in the X system will be 3 K.
Step 2: Using the obtained factor of change of temperature, find the specific heat of the material in the S.I. system.
The specific heat of the material in the X system is given to be $1400{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{X}}^{ - 1}}$ .
The factor by which the temperature changes in the X system is obtained as $\delta \left( {{t_x}} \right) = 3$
So, the specific heat of the material in the S.I. unit can be obtained by $1400 \times 3 = 4200{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$
Hence the correct option is A.
Note:Specific heat of a material refers to the amount of heat required to raise the temperature of a material of mass of $1{\text{kg}}$ by $1{\text{K}}$. Since $\delta \left( {{t_x}} \right) = 3$, in the new system, the specific heat of the material will be the amount of heat required to change the temperature of the material of mass of $1{\text{kg}}$ by ${\text{3K}}$.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

