(III) Twenty kilograms of granite quartz inclusions) has specific heat capacity

Question

(III) Twenty kilograms of granite quartz inclusions) has specific heat capacity Ceranite800J/(kg c) constitute the "hot rock" part of a sauna. Granite (with , and emissivity egr 0.75. The walls of the sauna are lined with red cedar. Red cedar has specific heat capacity C1500 J/(kg-c), and emissivity ecedar 0.33. The effective surface area of the rocks is 0.5m2, while that of the interior of the sauna is 24 a". The Stefan-Boltzmann constant is ? s: 5.67 x10-8 w/(m2-K4). Outside, it is wintertime, mnganite is inslated from contact with the walls so as to not ant them on fire] are at r,,-27c. The wall temperature remains effectively constant over the timescale of this question. (i) Compute the rate at which the rocks are radiating heat. (i) Compute the rate at which the rocks are absorbing heat radiation from the walls. (ii) Compute the NET rate at which the rocks are radiating heat. (b) Roughly how long does it take for the temperature of the granite rocks to reach 225 c? HINT: Feel fre to make resonable approxcimatione (e) At the instant that the temperature of the granite is 225 c, 200g of water, at an initial temperature of 30C, are splashed upon the rocks. The specific heat capacity and latent heat of vaporisation are d ??= 4 186J/(kgC) and Liao ~ m6kJ/kg respectively. The water is very quickly ["instantaneously"] heated to 100c, vaporises, and spreads throughout the sauna as steam at 100 c (1) How much energy is absorbed by the water as it undergoes its transformation to steam? (i) What is the temperature of the rocks immediately after the water boils away? me-scale Neglect the effects of therranl radiation on this short t Page 2 of4

Explanation / Answer

given

Granite:

M = 20 kg

Cg = 800 J/kg C

eg = 0.75

A = 0.5 m^2

Cedar

Cc = 1500 J/kg C

ec = 0.33

Ac = 24 m^2

sigma = 5.67*10^-8 W/m^2 *k^4

a. ti = 227 C

tw = 27 C

i. intiial rate of heat radiation by the rock = R1

R1 = sigma*eg*A*ti^4

t1 = 227 + 273.16 K

R1 = 1330.608066654 W

similiarly

ii. rate at which rocks are absorbing heat from the walls

R2 = sigma*ec*A*tw^4 = 75.94134 W

iii. net rate at which rock is heating = R = R1 - R2 = 1254.6667 W

b) for t = 225 C

this is a small fall in temeprature, hence we can assume the rat of heat dissipation to remain fairly constnat

hence

1254.6667*t = M*Cg*(227 - 225)

t = 2.5504781469054689 s

c) m = 200 g water

Tw = 30 C

Cw = 4186 J/ Kg C

Lw = 2256 kJ/kg C

i. energy absorbed by water = H

H = mCw(100 - Tw) + mLw

H = 509804 J

ii. temperature of rock when water boils away = t

MCg(225 - t) = H

t = -93.627 C = 179.5325 K

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.