A copper bowl of mass m 1 contains a mass m 2 of water, both at temperature T (w

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 masses 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. Assume the masses are measured in grams, temperatures in °C, specific heates in cal/g · °C, and the heat of vaporization of water in cal/g.)

Explanation / Answer

Use the formula mct.

Heat gained by bowl = m1 *0.0923* [100 - T]

Heat gained by 180 gm of water = m2 *1*[100-T]

Heat used to convert m4 g of water into steam = mL

= 539*m4 cal.

Total calories gained by water and bowl= (m1 *0.0923* [100 - T])+ (m2 *1*[100-T])+(539*m4)

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Heat lost by m3g of copper is m3* 0.0923*[θ - 100] cal.

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Equating the two

m3* 0.0923*[θ - 100] = (m1 *0.0923* [100 - T])+ (m2 *1*[100-T])+(539*m4)

θ=

--------------------------------------…

a) (m2 *1*[100-T])+(539*m4)

b) m2 *1*[100-T]

c) θ C.

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