A125g 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 data:

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

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explanation: