Question

a/ Compute the quantity of heat released by 25.0 g of steam initially at 100.0oC, when it is cooled to 34.0°C and by 25.0 g of water initially at 100.0oC, when it is cooled to 34.0°C.b/ Suppose you put your hand into this steam or this water (it is not recommended!), in which case your hand will be burnt more severely?Explain.The heat of fusion of water is 2.256×106J/kg and the specific heat of water is 4.19×103J/kg.K

Answer 1

When put into steam

Step-by-step explanation:

When a certain amount of steam at boiling temperature condenses (turning into water), the amount of heat released is

`Q_1 = m\lambda_v`

where in this case

m = 25.0 g = 0.025 kg is the mass of steam at 100.0°C

`\lambda_v=2.256\cdot 10^6 J/kg` is the latent heat of vaporization of water

So,

`Q_1=(0.025)(2.256\cdot 10^6)=56400 J`

Instead, the amount of heat released when the water at 100.0°C is cooled down to 34.0°C is given by

`Q_2=mC\Delta T`

where

m = 25.0 g = 0.025 kg is the mass of water

`C=4.19\cdot 10^4 J/kg K` is the specific heat of water

`\Delta T=100-34=66^(\circ)C` is the change in temperature

Therefore,

`Q_2=(0.025)(4.19\cdot 10^3)(66)=6913 J`

Since
`Q_1>Q_2`, we can say that your hand will burn more in the first case.