A copper bowl of mass m1 contains a mass m2 of water, both at temperature T (whi

Question

A copper bowl of mass m1 contains a mass m2 of water, both at temperature T (which is below the boiling point of water). A very hot copper cylinder with mass m3 is dropped into the water, causing the water to boil, with a mass m4 being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (Use any variable or symbol stated above along with the following as necessary: cw for the specific heat of water, cc for the specific heat of copper, and Lv for the latent heat of vaporization of water. Note that the c, w, and v are lowercase. Assume the masses are measured in grams, temperatures in °C, specific heats in cal/g · °C, and the heat of vaporization of water in cal/g.) (a) How much energy is transferred to the water as heat? Qwater = c (b) How much to the bowl? Qbowl = c (c) What is the original temperature of the cylinder? Ti = c

Explanation / Answer

Use the formula mct.
a.)Heat gained by bowl = m1 *0.0923* [100 - T] cal.
Heat gained by m2 gm of water = m1 *1*[100-T] cal.
Heat used to convert m4 of water into steam = m4L cal

Total calories gained by water and bowl= (m1 *0.0923* [100 - T] + m1 *1*[100-T]+ m4L )cal.
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b.)

Heat lost by m3 of copper is m3* 0.0923*[ - 100]
= m3*0.0923[ - 100] cal.
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c.) Equating the two
m3*0.0923[ - 100] cal = (m1 *0.0923* [100 - T] + m1 *1*[100-T]+ m4L )cal
m3*0.0923[ - 100] - m1 *0.0923*100 - m1 *1*100 - m4L = - m1 *0.0923*T - m1T

T = m3*0.0923[ - 100] - m1 *0.0923*100 - m1 *1*100 - m4L / (m1*0.0923 - m1)

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