A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the specific heat of the metal (0.900 j/g degrees Celsius), fund the final temperature of the block and the water

Final temperature is 34.2 °C

Explanation:

Given dа ta:

mass of metal = 125 g

temperature of metal = 93.2 °C

mass of water v = 100 g

temperature of water = 18.3 °C

specific heat of meta is = 0.900 j/g. °C

specific heat of water is = 4.186 j/g. °C

final temperature of water and metal = ?

Solution:

Q = m. c. ΔT

ΔT = T2-T1

now we will put the values in equation

Q1 = m. c. ΔT

Q1 = 125 g. 0.900 j/g. °C. 93.2°C - T2

Q1 = 112.5 (93.2°C - T2)

Q1 = 10,485 - 112.5T2

Q2 = m. c. ΔT

Q2 = 100. 4.186. (T2 - 18.3)

Q2 = 418.6. (T2 - 18.3)

Q2 = 418.6T2 - 7660.38

10,485 - 112.5T2 = 418.6T2 - 7660.38

10,485 + 7660.38 = 418.6T2 + 112.5T2

18145.38 = 531.1 T2

T2 = 18145.38/531.1

T2 = 34.2 °C