Question

Suppose the masses of the calorimeter, the water in it and the hot object made up of copper which is put in the calorimeter are the same. The initial temperature of the calorimeter and water is 30 0C and that of the hot object is 60 0C. The specific heats of copper and water are 0.09 cal / (gm 0C) and 1 cal / (gm 0C) respectively. What will be the final temperature of water?

Answer 1

Answer:

The initial temperature of the calorimeter and water is 30°C ... calorimeter and hot object both are made up of copper. ... (1). Heat loss by hot object = m×0.09×(60 ...

You visited this page on 12/8/19.

Answer 2

Answer:

32.3 °C

Explanation:

Heat gained by water and calorimeter = heat gained by the hot object.

Let the mass of each water, calorimeter and hot object be = m

Let the final temperature be = T

Thus,

Heat gain by water and calorimeter

= m×0.09×(T-30) + m×1×(T-30)  --- 1

 Heat loss by the hot object

= m×0.09×(60-T)  --- 2

 By equating (1) and (2), we will get,

= (T-30) ×1.09 = (60-T) ×0.09  .

 On solving for T we will get  T = 32.3C

Therefore, the final temperature of water is 32.3°C