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IRISSAK [1]
3 years ago
15

Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your s

kin from steam as opposed to hot water at the same temperature.Assume that water and steam, initially at 100?C, are cooled down to skin temperature, 34?C, when they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a constant specific heat capacity c=4190J/(kg?K) for both liquid water and steam.How much heat H1 is transferred to the skin by 25.0 g of steam onto the skin? The latent heat of vaporization for steam is L=2.256
Chemistry
1 answer:
Ulleksa [173]3 years ago
3 0

Answer:

H1 = 63.3 kJ

Explanation:

When the steam at 100C contacts your skin, it is first condensed to water at 100C. The heat involved with this change in state is calculated using the latent heat of vaporization:

Q = ML = (25.0 g)(2.256 kJ/g) = 56.4 kJ

The water at 100C is then cooled to 34C. The heat involved with this temperature change is calculated using the specific heat capacity. (Because a temperature <em>change </em>is used here, it is not necessary to convert °C to K in order for the units to properly cancel)

Q = mcΔt = (25.0 g)(1kg/1000g)(4190Jkg⁻¹K⁻¹)(100C - 34C) = 6913.5 J = 6.9135 kJ

The total heat H1 that is transferred to the skin is the sum of the two heat values above:

H1 = (56.4 kJ) + (6.9135 kJ) = 63.3 kJ

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So I saw this question: If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + Na
Nataly_w [17]

mass of PbI₂ = 27.6606 g

<h3>Further explanation</h3>

Given

Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃

28.0 grams of Pb(NO₃)₂ react with 18.0 grams of NaI

Required

mass of PbI₂

Solution

Balanced equation

Pb(NO₃)₂ + 2NaI → PbI₂ + 2NaNO₃

The principle of a balanced reaction is the number of atoms in the reactants = the number of atoms in the product

mol Pb(NO₃)₂ :

= 28 : 331,2 g/mol

= 0.0845

mol NaI :

= 18 : 149,89 g/mol

= 0.12

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Pb(NO₃)₂ : 0.0845 : 1 = 0.0845

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