Question

Calculate the heat released when 68.0 gram of steam at 124⁰C is converted to water

    at 45⁰C

Answer 1

released can be calculated by the following formula.

$q=ms\Delta t$

m = Mass of sample

s = Specific heat of steam

$\Delta t$ =Temperature change

From the given,

Mass of sample = 68.0 g

Specific heat of steam = 1.99 J/(g°C)

$\Delta t=124-100=24°C$

$q=68x1.99x24=3247 J$

s₁ - Specific heat of water, 4.184 J

$\Delta t=100-45=55°C$

$q_1=68x4.184x55=$ 15648

Q=q+q₁=3247+15648=18895 J