What mass of steam initially at 130 degree C is needed to warm 200 g of water in

Question

What mass of steam initially at 130 degree C is needed to warm 200 g of water in a 100-g glass container from 17.0 degree C to 47.0 degree C? Solve it Conceptualize Imagine placing water and steam together in a closed insulated container. The system eventually reaches a uniform state of water with a final temperature of 47.0 degree C. Categorize Based on our conceptualization of this situation, we categorize this example as one involving calorimetry in which a phase change occurs. The calorimeter is an isolated system for energy: energy transfers between the components of the system but does not cross the boundary between the system and the environment. Write an equation to describe the calorimetry process: (1) Q cold = Q hot The steam undergoes three processes: a decrease in temperature to 100 degree C, condensation into liquid water, and finally a decrease in temperature of the water to 47.0 degree C. Find the energy transfer in the first process using the unknown mass ms of the steam: Q1 = mscs Delta Ts Find the energy transfer in the second process: Q2 = Lvms = Lv(0 - ms) = -msLv Find the energy transfer in the third process: Q3 = mscw Delta T hot water Add the energy transfers in these three stages: Q hot = Q1 + Q2 + Q3 = ms(cs Delta Ts - L v + cw Delta T hot water) Qcold = mwcw Delta T cold water + mgcg Delta T glass The 17.0 degree C water and the glass undergo only one process, an increase in temperature to 47.0 degree C. Find the energy transfer in this process Substitute Equations (2) and (3) into Equation (1): If we instead used 57.0 g of water, how much 100 degree C steam remains in thermal equilibrium? 8.13

Explanation / Answer

ms = -[(0.057*4186*(47-17))+(0.1*837*(47-17)] / [-(2.26*10^6)+(4186*(100-47)] = 3.89 g

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