150 g of water at 38 o C were mixed with 120 g of cold water in a calorimeter. W

Question

150 g of water at 38 oC were mixed with 120 g of cold water in a calorimeter. When thermal equilibrium was established, the temperature of the mixture was measured at 29 oC. The specific heat of water is 1.00 cal/ g oC. The calorimeter and the 120 g of water absorbed ______cal.

Enter your answer with 2 sig fig.

150 g of water at 38 °C were mixed with 120 g of cold water in a cup calorimeter When thermal equilibrium was established, the temperature of the mixture was measured at 29°C. The specific heat of water is 1.00 cal g oC. Part A Which of the following statements is true? Choose all that apply. O he hot water absorbs heat from the calorimeter O he hot water releases heat. O he calorimeter releases heat, but the cold water absorbs heat O The cold water absorbs heat O The calorimeter absorbs heat.

Explanation / Answer

Heat absorbed by calorimeter+120g water=heat lost by 150g hot water at 38 degree Celsius.

Heat change=ms*temperature difference=150*1*9=1350cal

Also we know that 1cal=4.18J

So heat change=1360*4.18=5643J=5.643KJ=5.6KJ

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