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

Answer:

556.5 KJ

Explanation:

We are given that

Mass of water=m=2.5 kg=$2.5\times 1000=2500 g$

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

$Q=2500\times 4.184\times 53.2$

$Q=556472 J$

1 KJ=1000 J

$Q=\frac{556472}{1000}=556.472\approx 556.5 KJ$

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

Answer 2

Answer:

556.5 J

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.