A 125G sample of water was heated to 100.0°C and then I borrow platinum 20.0°C is dropped into the beaker the temperature of the platinum in the beaker quickly rose 235.0°C the specific heat of platinum is 0.13 j/g°C. The specific heat of water is 4.184 J/g°C. What is the mass of platinum

mass of platinum = 2526.12 g

Explanation:

Given dа ta:

Mass of water = 125 g

Initial temperature of water = 100.0°C

Initial temperature of Pt = 20.0°C

Final temperature = 235°C

Specific heat of Pt = 0.13 j/g°C

Specific heat of water = 4.184 j/g°C

Mass of platinum = ?

Solution:

Formula:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

Q (w) = Q (Pt)

m. c. (T2 - T1) = m. c. (T2 - T1)

125 g * 4.184 j/g°C * (235°C - 100.0°C) = m * 0.13 j/g°C * (235°C - 20°C)

125 g * 4.184 j/g°C * 135°C = m * 0.13 j/g°C * 215°C

70605 j = m*27.95 j/g

m = 70605 j / 27.95 j/g

m = 2526.12 g