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saw5 [17]
2 years ago
10

Your observation and assessment data should help you _______ your curriculum A. change B. modify and individualize C. evaluate t

he effectiveness of D. complement
Chemistry
2 answers:
Degger [83]2 years ago
8 0
The answer is b. Modify and individualize
joja [24]2 years ago
4 0
Hello,
I believe the answer is C. A and B say that it would help to change it but when we think about it data and opinions are not what people use to change a curriculum or program. Mainly they use surveys, which they didn't use here. Option D doesn't make any sense, data doesn't complement a program it supports some sort of theory which leads us to C. C is the only option where they use the data for effectiveness. Hope this helps!
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xxTIMURxx [149]

Answer:

T_2=17.8\°C

Explanation:

Hello,

In this case, we can solve this problem by noticing that the heat lost by the warm water is gained by the ice in order to melt it:

Q_{water}=-Q_{ice}

In such a way, the cooling of water corresponds to specific heat and the melting of ice to sensible heat and specific heat also that could be represented as follows:

m_{water}Cp_{water}(T_2-T_{water})=-m_{ice}\Delta H_{melting,ice}-m_{ice}\Cp_{ice}(T_2-T_{ice})

Thus, specific heat of water is 4.18 J/g°C, heat of melting is 334 J/g and specific heat of ice is 2.04 J/g°C, thus, we can compute the final temperature as shown below:

m_{water}Cp_{water}(T_2-T_{water})+m_{ice}Cp_{ice}(T_2-T_{ice})=-m_{ice}\Delta H_{melting,ice}\\\\T_2=\frac{-m_{ice}\Delta H_{melting,ice}+m_{water}Cp_{water}T_{water}+m_{ice}Cp_{ice}T_{ice}}{m_{water}Cp_{water}+m_{ice}Cp_{ice}} \\\\T_2=\frac{-50.0*334+100*4.18*67+50.0*2.04*-20.0}{100*4.18+50.0*2.04} \\\\T_2=17.8\°C

Best regards.

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3 years ago
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