1. A 5.88 kg piece of granite with a specific heat of 0.803 J g -1 °C -1 and a t

Question

1. A 5.88 kg piece of granite with a specific heat of 0.803 J g-1 °C-1 and a temperature of 85.1 °C is placed into 2.00 L of water at 19.0 °C. When the granite and water come to the same temperature, what will the temperature be?

2. The combustion of methane (the chief component of natural gas) follows the equation:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

H° for this reaction is -802.3 kJ. How many grams of methane must be burned to provide enough heat to raise the temperature of 202.3 mL of water from 24.09 °C to 42.67 °C?

3. How much heat, in joules and in calories, must be removed from 1.58 mol of water to lower its temperature from 26.6 to 14.1°C? Include the sign in your answer.

4. A vat of 4.07 kg of water underwent a decrease in temperature from 61.38 to 58.44 °C. How much energy in kilojoules left the water? Include the sign in your answer. (For this range of temperature, use a value of 4.18 J·g-1·°C-1 for the specific heat of water.)

Explanation / Answer

1.

Heat lost by Granite= Heat gained by water

Mass of granite* specific heat of granite* (85.1-T)= Mass of water* specific heat of water*(T- 19)

T= Equilibrium temperature reached by water and granite.

Assuming water density to be 1g/cc , 2 L of water correspond to 2000 gms

Mass of water= 2000gm, Specific heat of water= 4.18 j/g.deg.c

5.88*0.803*(85.1-T)= 2000* 4.18*(T-19)

4018.12-4.72T= 8360T-158840

4018.12+158840= 8364.72T

T=19.47 deg.c

2. delH= Enthalpy change of water = mass of water* specific heat of water* temperature differecne

Assuming density of water to 1g/ml, specific heat of water= 4.18 j/g.deg.c

Enthalpy change= 202.3*1*4.18*(42.67-24.09)=15711.51 joules= 15.712 Kj

Enthalpy change has to come from combustion of methane which is 802.3 KJ/mol of methane

Moles of methnane required= 15.712/802.3=0.019584

Mass of methane =molecular weight* moles= 0.019584*16=0.313 gms

3. Heat to be removed= moles of water* mass of water* specific heat of water* Temperature difference= 1.58*18*4.18*(26.6-14.1)=1486 joules

4. Heat left = 4.07*1000*4.18*(61.38-58.44)=50017.04 joules

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