Question

A What is the amount of heat entering your skin when it receives the heat released by 25.1 g of steam initially at 100 ∘C that cools to 34.1 ∘C? Assume Lv=2256×103J/kg and cw=4190J/(kg⋅K). Q = J SubmitGive Up Part B What is the amount of heat entering your skin when it receives the heat released by 25.1 g of water initially at 100 ∘C that cools to 34.1 ∘C? Assume cw=4190J/(kg⋅K). Q = J

Answer 1

Given that :

mass of the steam, m_s = 25.1 g = 0.0251 kg

initial temperature, T₁ = 100 ⁰C = 373.15 K

final temperature, T₂ = 34.1 ⁰C = 307.25 K

latent heat of vaporization, $\Delta$L_v = 2256 x 10³ J/kg

specific heat constant of water, C_w = 4190 J.kg.K

(A) The amount of heat entering skin when it receives the heat released which is given as :

$\Delta$Q = m_s$\Delta$L_v + m_s C_w$\Delta$T    { eq.1 }

where, $\Delta$T = change in temperature = T₁ - T₂

inserting the values in above eq.

$\Delta$Q = (0.0251 kg) (2256 x 10³ J/kg) + (0.0251 kg) (4190 J/kg.K) [(373.15 K) - (307.25 K)]

$\Delta$Q = (56625.6 J) + (105.17 J/K) (65.9 K)

$\Delta$Q = (56625.6 J) + (6930.7 J)

$\Delta$Q = 63556.3 J

(b) The amount of heat entering skin when it receives the heat released which is given as :

$\Delta$Q = m_w C_w$\Delta$T    { eq.2 }

where, m_w = mass of water = 0.0251 kg

inserting the values in eq.2,

$\Delta$Q = (0.0251 kg) (4190 J/kg.K) [(373.15 K) - (307.25 K)]

$\Delta$Q = (105.17 J/K) (65.9 K)

$\Delta$Q = 6930.7 J