physic: heat and internal?

★ 題目應該打錯了一個「單位 」: 不是 "80°C", 而是 "80g"
∴ 題目是 : adding 200g of water at 90°C to 80g of noodles at 20°C
# [ 這是我的推斷 ,請對答案 , 便知是否正碓 ]
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Solution :-

Let t = temp. of noodles after adding water,
heat gained from noodles = heat lost from water
　　　2000 x (200/1000) x (t-20) = 4200 x (80/1000) x (90-t)
　　　　　　　　　400 x (t-20) = 336 x (90-t)
　　　　　　　　　400t - 8000 = 30240 - 336t
　　　　　　　　　　　　736t = 38240
　　　　　　　　　　　　∴ t = 52.0 °C (to 1 dec. place)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ COMPLETED ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note :-
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(1) Formula: Q = C x m x ΔT
(2) specific heat capacity of water = 4200 J Kg ⁻¹ °C⁻¹
( can be obtained from the TOP / BOTTOM of this exercise, or in the APPENDIX at the end of your book. )