Question

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?

Answer 1

T2 = 24.56°C

Step-by-step 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

Answer 2

Answer: Final temperature for given mass of water is
`24.6^o C`.

Step-by-step explanation:

Given: mass = 630.1 g, specific heat = 418
`J/g^(o) C`, q = 950 J, T =
`24.2^o C`

Formula used:
`q = mC \Delta T`

where, q = heat energy

m = mass

C = specific heat

`\Delta T` = change in temperature =
`(T_2 - T_1)`

950 J =
`630.1 g * 4.18 J/g^o C * (T_2 - 24.2^o)`

`T_2 = 24.6^o C`