Question

Determine the Kelvin temperature for each of the following:

(a) human body temperature;

(b) the boiling point of water(at the standard pressure of 1 atm);

(c) the coldest day u can remember;

(d) the boiling point of liquid nitrogen(-196$°$C);

(e) the melting point of lead(327$°$C)

Short answer

$$\begin{gathered} \text{(a)}310.15\text{K} \\ \text{(b)}373.15\text{K} \\ \text{(c)}277.15\text{K} \\ \text{(d)}77.15\text{K} \\ \text{(e)}600.15\text{K} \end{gathered}$$

Step-by-step solution

Part (a): Step 1. Given information

The human body temperature in Celsius scale is $37°\text{C}$.

Part (a) Step 2: Calculation

The relation between the temperature in Kelvin scale and that in Celsius scale is given by

$K=C+273.15...................\left(1\right)$

Here, $K$is the temperature in Kelvin scale and $C$is that in Celsius scale.

Substitute $37$for $C$into equation (1) to calculate the required value.

$\begin{array}{rcl}K& =& 37+273.15\\ & =& 310.15\text{K}\end{array}$

Part (b) Step 1: Given information

The boiling point of water at standard pressure is $100°\text{C}$.

Part (b): Step 2. Calculation

Substitute $100$for $C$into equation (1) to calculate the required value.

$\begin{array}{rcl}K& =& 100+273.15\\ & =& 373.15\text{K}\end{array}$

Part (c): Step 1. Given information

The coldest day that can be remembered is $4°\text{C}$.

Part (c): Step 2. Calculation

Substitute $4$for $C$into equation (1) to calculate the required value.

$\begin{array}{rcl}K& =& 4+273.15\\ & =& 277.15\text{K}\end{array}$

Part (d): Step 1. Given information

The boiling point of liquid nitrogen is $-196°\text{C}$.

Part (d): Step 2. Calculation

Substitute $-196$for $C$into equation (1) to calculate the required value.

$\begin{array}{rcl}K& =& -196+273.15\\ & =& 77.15\text{K}\end{array}$

Part (e) : Step 1. Given information

The melting point of lead is $327°\text{C}$.

Part (e): Step 2. Calculation

Substitute $327$for $C$into equation (1) to calculate the required value.

$\begin{array}{rcl}K& =& 327+273.15\\ & =& 600.15\text{K}\end{array}$

Final answer

The required temperatures are as follows: