Question

level 20
this table shows the temperature of a liquid as it cooled, after being heated.
time (s) | temperature (°c)
0 | 95
20 | 83
40 | 71
60 | 59
80 | 47
100 | 35
120 | 23
what does the initial value and rate of change mean in this situation?
select an answer
a the liquid was heated to 120 °c and cooled by 20 °c every second.
b the liquid was heated to 95 °c and cooled by 0.6 °c every second.
c the liquid was heated to 95 °c and cooled by 12 °c every second.
d the liquid was heated to 95 °c and cooled by 6 °c every second.

Explanation:

Step1: Find initial temperature

Initial time is 0s, so initial temp = 95°C.

Step2: Calculate time interval

Time increases by 20s each step (20-0=20, 40-20=20, etc.).

Step3: Calculate temp change per interval

Temp change: 83-95=-12°C, 71-83=-12°C, etc. Rate = $\frac{-12}{20}=-0.6$°C/s.

Answer:

B. The liquid was heated to 95 °C and cooled by 0.6 °C every second.