Question

The temperature of 600 g of cold water rises by 15° C when 300 g of hot water at 50° C is added to it. What was the initial temperature of the cold water?

Answer 1

Mass of cold water = 600 g

Mass of hot water = 300 g

Temperature of hot water = 50° C

Let initial temperature of cold water be t_i

Let the final temperature be t

Gain in temperature of cold water = (t - t_i) = 15° C

Loss of heat from hot water = (50 - t)

Heat energy given by hot water = mc△t

= 300 × 4.2 x (50 – t) [Equation 1]

Heat energy taken by cold water = 600 x 4.2 x (t - t_i) [Equation 2]

Assuming that there is no loss of heat energy

Heat energy given by hot water = heat energy taken by cold water

Equating equations 1 & 2, we get,

300 × 4.2 × (50 − t) = 600 × 4.2 × 15

⇒ 3 × (50 − t) = 6 ×1 5

⇒ 50 − t = 2 ×15

⇒ 50 −t = 30

⇒ t = 50 − 30 = 20°C

final temperature (t) = 20°C

initial temperature = ?

We know,

t - t_i = 15

20 - t_i = 15

Therefore, t_i = 20 - 15 = 5° C

Hence, initial temperature = 5° C