Question

please answer the question, thank you.

W2 Trial 2 43.999 درهم العفن علمنعم دقة 20.0g Trial L mass of the dry metal 43.989 70.09 compas.ct 90.0°c ene . Co wake 26.6°C final temp (water and metal) 27.4°C Tempecent changed at binal -62.6°C temyexe Chonge LAT) oktober 0.8°C 90.2 26.900 30.1°C -60.1°C 3.2°C Specielrout of metal in Trial = 0.8319°C specific reatot retal in Trial 2 = 0.3s J19°C - Average specific reat = 0.21Jlgoc - Metal : Rb • Density = 1.63 g/m2 QUESTION: A colorimeter contains 110. Od water at some What is Teinen it a 41.00 g sample of your retal at 355.0°C is added? vse the average syelikte heet.

Answer 1

In the above experiment, heat is lost by the metal while heat is gained by water (assuming no other losses).

Therefore, heat lost by the metal = heat gained by water

Heat lost by the metal = - m_metalC_metal(T_final - T_{initial, metal})

Heat gained by water = m_waterC_water(T_final-T_{initial, water})

=> - m_metalC_metal(T_final - T_{initial, metal}) = m_waterC_water(T_final-T_{initial, water}) ---(1)

Given:

m_metal = 41 g

C_metal = 0.21 J/g/^oC

T_{initial, metal} = 355^oC

m_water = 110 g (since density of water = 1g/cc)

C_water = 4.184 J/g/^oC

T_{initial, water} = 5^oC

Sunbstituting the above values in equation (1), we get

- 41 * 0.21 * (x - 355) = 110 * 4.184 *(x - 5)

=> -8.61 x + 3056.55 = 460.24x - 2301.2

=> (460.24 + 8.61)x = 3056.55+2301.2

=> 468.85 x = 5356.75

=> x = 5357.75/468.85

=> x = 11.43^oC

So, the final temperature of the water = final temperature of metal = 11.43^oC