Hot water is added to three times its mass of water at 10oC and the resultant temperature is 20oC. What is the initial temperature of the hot water?

austinaa514

2 years ago

The answer is supposed to be 80 how did you get 50

Emzyuniques7

2 years ago

i no understand ooo

mayor177

9 months ago

Ans not there

Maximum2025

7 months ago

MCT=MCT
ratio of the masses=1:3
Therefore
1CT=3CT
heat change
From 10c to 20c=+10
the loss must be the same value but with opposite signs
3CT lost -10c as 10+(-10)=0(heat lost is equal to heat gained.
back to the formula
1CT=3CT
1(10)c=3(-10)c
10c=-30c
divide both sides by c to cancel it out(c is constant)
10c/c=-30c/c
10°c gained(positive sign)= -30°c lost(negative sign)
10°c to 20°c gained, 30°c lost from 50°c to 20°c.

austinaa514

2 years ago

The place 10 degrees was used is supposed to be 100 degrees

YOUNGPHARM

10 years ago

The answer is wrong bcosNCOs 20°c is the resultant temperature which must be less than the temperature of the hot water..

It is calculated as follows;

m×c t = m×c×t (heat loss by hot water equall heat gain b water @ 10°c)

3m ×c×(t-20)=m×c×(20-10)

3×(t-20)=10

t-20=10÷3

t=3.3333+20=23.33°c

Miracles1237

3 years ago

3m×c(-20)=m×c(20-10)
3-60=10
3=10+60
3=70
=23.33°c

kayceematts

4 years ago

I think the answer is D

Initial temp.=t°c
Mass of hot water=mg
Mass of cold water=10°c
Final temp.=20°c

Rise in temp.of cold water=(20-10)°c
Rise in temp.of hot water=(t-20)°c

Using the principle of mixtures
Heat energy by hot water=Heat energy by cold water.
Given as:
m×(t-20)=3m×(20-10)
(t-20)=2m×(20-10)
t-20c=40m-20m
t-20c=20m
Divide bother sides by c
t-20=20
t=40°c