Question

9. A mass of 50 g of a certain metal at 150°C is
immersed in 100 g of water at 11°C. The final
temperature is 20°C. Calculate the specific heat
capacity of the metal. Assume that the specific heat
capacity of water is 4.2 J g-1 K-1
Ans. 0.582 J g-1 K-1​

Answer 1

Answer:

Given

$mass \: of \: the \: metal \: = 50g \\ initial \: temperature \: of \: metal \\ = {150}^{0}c = 150 + 273 = 423 k \\ mass \: of \: water = 100g \\ initial \: temperature \: of \: water \\ = {11}^{0} c = 11 + 273 = 284k \\ final \: temperature = {20}^{0} c = 293k$

According to law of method of mixtures

Heat lost by hotter body =Heat gained by cooler body

mass of metal x specific heat of metal x

change in temperature of metal

=

mass of water x specific heat of water x

Change in temperature of water

$50 \times s \times (423 - 293) \\ = \\ 100 \times 4.2 \times (293 - 284) \\ s = \frac{420 \times 9}{50 \times 130 } \\ specific \: heat \: of \: metal \: \\ = 0.5815 \: joule \: {gram}^{ - 1} {kelvin}^{ - 1}$