Maharashtra Class 10 Answered
Suppose the masses of calorimeter, th 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°C and that of hot object is 60°C. The specific heat of copper and water are 0.09 cal/gm°C and 1 cal/gm°C respectively. Also the specific heat of calorimeter is same as copper. What will be the final temperature of water (mixture).
Asked by Anshul | 16 Mar, 2019, 09:36: AM
Expert Answer
Heat gain by water and calorimeter = heat gain by hot object.
let m (in gram ) be mass of each i.e., water, calorimeter and hot object.
calorimeter and hot object both are made up of copper.
Let T be the final temperature.
Heat gain by water and calorimeter = m×0.09×(T-30) + m×1×(T-30) ........................(1)
Heat loss by hot object = m×0.09×(60-T) .......................................(2)
By equating (1) and (2), we get, (T-30)×1.09 = (60-T)×0.09 ...............(3)
solving for T from eqn.(3), we get T = 32.3 °C
Answered by Thiyagarajan K | 16 Mar, 2019, 03:06: PM
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