Question

Calculate the amount of energy required (in joules) to heat 2.5 kg of water from 16.7 °C to 69.9 °C. The specific heat of water is 4.184 J/g °C.

Answer 1

556.5 J

Step-by-step explanation:

mass of water, m = 2.5 kg

Initial temperature of water, T1 = 16.7 °C

final temperature of water, T2 = 69.9 °C

Specific heat of water,c = 4.184 J/g °C

The amount of energy required to rise the temperature is given by

H = m x c x ΔT

where, ΔT is the rise in temperature.

H = 2.5 x 4.184 x 1000 x ( 69.9 - 16.7)

H = 556.5 x 1000 J

H = 556.5 kJ

Thus, the amount of heat required is 556.5 J.

Answer 2

556.5 KJ

Step-by-step explanation:

We are given that

Mass of water=m=2.5 kg=
`2.5* 1000=2500 g`

1 kg=1000 g

`T_1=16.7^(\circ)`

`T_2=69.9^(\circ)`

`\Delta T=T_2-T_1=69.9-16.7=53.2^(\circ)`C

Specific heat of water=
`c=4.184J/g^(\circ)C`

We have to find the amount of energy required .

We know that

`Q=mc\Delta T`

Using the formula

`Q=2500* 4.184* 53.2`

`Q=556472 J`

1 KJ=1000 J

`Q=(556472)/(1000)=556.472\approx 556.5 KJ`

Hence, the amount of energy required to heat 2.5 kg water=556.5 KJ