# Please help me with physics question on temperature ?

In a calorimeter, where there are 200 g of water at a temperature of 20 ° C, 300 g of water is poured at a temperature of 100 ° C. What is the final temperature of the 500 g of water, assuming that heat losses are negligible?

### 5 Answers

- 1 month ago
Heat gained by 200 g of water = mc∆T = 200(1)(T-20)

Heat lost by 300 g of water = 300(1)(100 - T)

Heat lost = Heat gained

200(1)(T-20) = 300(1)(100 - T)

T = 68 deg C

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- PhilomelLv 71 month ago
200*20=4000

300*100=30000

34000/500=68 degC

- charlatanLv 71 month agoReport
old school

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- charlatanLv 71 month ago
(20*200+300*100)/(200+300)=68 deg C

using Calorie unit is simpler and easier

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- FiremanLv 71 month ago
Let the final temperature of the water is t*C, by the law of energy conservation:

=>Q(gain) by 200 gm water = Q(loss) by 300 gm of water

=>m(200)s∆t = m(300)s∆t

=>200 x (t - 20) = 300 x (100-t)

=>200t - 4000 = 30000 - 300t

=>500t = 34000

=>t = 34000/500 = 68*C

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- oubaasLv 71 month ago
4.186(200*20+300*100) = 4.186**Tf(200+300)

specific heat of water cross

Tf = (200*20+300*100) / 500 = 68°C

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