Question

580 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass is 900 grams and initial temperature 24°C (the specific heat capacity of aluminum is 0.9 J/K/gram). After a short time, what is the temperature of the water? °C What simplifying assumptions did you have to make? Assume negligible energy transfer between the surroundings and the system consisting of the pan plus the water. Assume that energy diffuses throughout the pan and water so that all parts reach the same final temperature. Assume that the water temperature hardly changes. Assume the specific heat capacity is nearly constant in this temperature range. t

Answer 1

Answer:

81.04°C

Explanation:

Heat loss by water = Heat gained by Aluminum

Heat loss by water;

H = MCΔT

ΔT = 100 -  T2

M = 580g

c = 4.2

H = 580 * 4.2 (100 - T2)

H =  243600  - 2436T2

Heat ganed by Aluminium

H = MCΔT

ΔT = T2 - 24

M = 900g

c = 0.9

H = 900 * 0.9 (T2 - 24)

H = 810 T2 - 19440

243600  - 2436T2 = 810 T2 - 19440

243600 + 19440 =  810 T2 + 2436T2

263040 = 3246 T2

T2 = 81.04°C

Assumption;

Assume that energy diffuses throughout the pan and water so that all parts reach the same final temperature.