For drinks during the day ,Peter has a cup of hot coffee at a temperature of 90 degree celsius and - Science - Gravitation

Question

For drinks during the day ,Peter has a cup of hot coffee at a
temperature of 90 degree celsius and a cup of cold water ,with a
minimum temperature of 5 degree celsius THe cups are of identical
type and size and volume of each drink is same Peter leave the cup
sitting in a room of temperature 20 degreee celsius
WHAT ARE THE TEMPERATURE OF COFFEE AND WATER LIKELY TO BE AFTER
10 MINUTES

70 degree and 10 degree
90 degree and 5 degree
70 degree and 25 degree
20 degree and 20 degree

EXPLAIN WITH REASON PLZ

Rajendra Kumar · Accepted Answer

From Newton's law of cooling, time taken to cool a liquid from temperature $T_{1}$ to $T_{2}$ with surrounding temperature is given as:

$t = \frac{1}{k} \ln (\frac{T_{2} - T_{0}}{T_{1} - T_{0}})$

For coffee to cool from $90^{ο} C$ to $T_{2}$ , time taken

$10 = \frac{1}{k} \ln (\frac{T_{2} - 20}{90 - 20}) = \frac{1}{k} \ln (\frac{T_{2} - 20}{70}) ...... (1)$

For coffee at $5^{ο}$ , time taken to raise the temperature,

$10 = \frac{1}{k} \ln (\frac{T_{2}^{'} - 20}{5 - 20}) = \frac{1}{k} \ln (\frac{T_{2}^{'} - 20}{- 15}) ...... (2)$

Dividing eq. (1) by eq. (2)

$\frac{\ln (\frac{T_{2} - 20}{70})}{\ln (\frac{T_{2}^{'} - 20}{- 15})} = 1 \Rightarrow (\frac{T_{2} - 20}{70}) = (\frac{T_{2}^{'} - 20}{- 15}) \Rightarrow - 15 T_{2} + 300 = 70 T_{2}^{'} - 1400 \Rightarrow 70 T_{2}^{'} + 15 T_{2} = 1400$

This is the relation between final temperature of the coffee at two different initial temperature. This relation is satisfied for only $T_{2}^{'} = 20^{ο} C a n d T_{2} = 20^{ο} C .$ So the last option is correct.