Question

1.If 245 Joules is required to change the temperature of 14.4 g of Chromium by 38.0 K, What is the specific heat of capacity of Chromium?

2. Calculate the energy in the form of heat (in kJ) required to convert 225 grams of liquid water at 21.0 °C
to steam at 115 °C. (Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities:
liquid water = 4.184 J/g·K, steam = 1.92 J/g·K)

Answer 1

Question 1

we can use the following formula

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat (units J/g∙°C), ∆ is a symbol meaning "the change in"

∆T = change in temperature (°C Celcius)

Q = 245 Joules m = 14.4 g ∆T = 38 K or 38 °C

Q = mc∆T

c = Q / m∆T

c = 245 Joules /14.4 g x 38 °C = 0.4477 J/g∙°C

Specific heat of capacity of Chromium is 0.4477 J/g∙°C

Question 2

we can use the same formula

Q = 225 g x  4.184 J/g∙°C x (100 ∙°C- 21 ∙°C) + 225 g x  2256 J/g + 225 g x 1.92 J/g∙°C x (115 ∙°C- 100∙°C)

Q = 74370.6 J + 5,07,600 J + 6480

Q =  5,88,450.6 Joules or 588.451 Kilo Joules

588.451 Kilo Joules of heat (in kJ) required to convert 225 grams of liquid water at 21.0 °C to steam at 115 °C.