What is the final temperature of 630.1 g of water (specific heat = 4.18 J/g.°C) at 24.2°C that absorbed 950. J of heat?

T2 = 24.56°C

Explanation:

Given data:

Mass of water = 630.1 g

Specific heat of water = 4.18 J/g.°C

Initial temperature = 24.2°C

Heat absorbed = 950 J

Final temperature = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

by putting values

950 J = 630.1 g × 4.18 J/g.°C ×  (T2 - 24.2°C)

950 J = 2633.82 j/°C ×  (T2 - 24.2°C)

950 J / 2633.82 j/°C =  (T2 - 24.2°C)

0.36°C  + 24.2°C = T2

T2 = 24.56°C

16

explanation:

the marks go by 4s