Question

How many grams of water can be heated from 25.0^oC to 36.7^oC by the heat released from 78.0 g of copper that cools from 88.5^oC to 32.8^oC?

How many grams of copper can be cooled from 79.7 to 26.0 °C by the heat gained by 100.0 g of water that has an increase in temperature from 21.0 to 39.8 °C?

A vat of 5.25 kg of water underwent a decrease in temperature from 60.33 to 58.21 °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.)

In the reaction between formic acid (HCHO₂) and sodium hydroxide, water and sodium formate (NaCHO₂) are formed. To determine the heat of reaction, 75 mL of 1.07 M HCHO₂ was placed in a coffee cup calorimeter at a temperature of 20.3 °C, and 45 mL of 1.78 M NaOH, also at 20.3 °C, was added. The mixture was stirred quickly with a thermometer, and its temperature rose to 24.1 °C. Write the balanced chemical equation for the reaction. Calculate the heat of reaction in joules. Assume that the specific heats of all solutions are 4.184 J g^-1°C^-1 and that all densities are 1.00 g mL^-1. What is the heat of reaction per mole of acid (in units of kJ mol^-1).

Write the balanced reaction that takes place:

Answer 1

1. How many grams of water can be heated from 25.0^oC to 36.7^oC by the heat released from 78.0 g of copper that cools from 88.5^oC to 32.8^oC?

Heat released by copper = mass of copper * specific heat capacity of copper * temperature change = 78.0 g * (0.385 J/ g ⁰C) * ( 32.8 - 88.5) ⁰ C = 1672.7 J

Heat gained by water = mass of water * specific heat capacity of water * temperature change = x g * (4.184 J/ g ⁰C) * ( 36.7 - 25) ⁰ C = 49x Joules

So, 49x = 1672.7

Or, x = 34.1

So, mass of water = 34.1 g

2. How many grams of copper can be cooled from 79.7 to 26.0 °C by the heat gained by 100.0 g of water that has an increase in temperature from 21.0 to 39.8 °C?

Heat gained by water = mass of water * specific heat capacity of water * temperature change = 100 g * (4.184 J/ g ⁰C) * ( 39.8 - 21) ⁰ C = 7866 Joules

Heat released by copper = mass of copper * specific heat capacity of copper * temperature change = x g * (0.385 J/ g ⁰C) * ( 26 - 79.7) ⁰ C = -20.7x Joules

So, 20.7x = 7866

Or, x = 380

So, mass of copper = 380 grams

3. A vat of 5.25 kg of water underwent a decrease in temperature from 60.33 to 58.21 °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.)

Mass of water = 5.25 kg = 5250 g

Heat released by water = mass of water * specific heat capacity of water * temperature change = 5250 g * (4.184 J/ g ⁰C) * ( 58.21 - 60.33) ⁰ C = - 46567 Joules = - 46.567 kJ

So, amount of heat released = 46.567 kJ

4. In the reaction between formic acid (HCHO₂) and sodium hydroxide, water and sodium formate (NaCHO₂) are formed. To determine the heat of reaction, 75 mL of 1.07 M HCHO₂ was placed in a coffee cup calorimeter at a temperature of 20.3 °C, and 45 mL of 1.78 M NaOH, also at 20.3 °C, was added. The mixture was stirred quickly with a thermometer, and its temperature rose to 24.1 °C. Write the balanced chemical equation for the reaction. Calculate the heat of reaction in joules. Assume that the specific heats of all solutions are 4.184 J g^-1°C^-1 and that all densities are 1.00 g mL^-1. What is the heat of reaction per mole of acid (in units of kJ mol^-1).

Balanced equation: HCOOH (aq) + NaOH (aq) = HCOONa (aq) + H₂O (l)

Mass of HCHO2 solution = volume * density = 75 mL * ( 1 g / mL) = 75 g

Mass of NaOH solution = volume * density = 45 mL * ( 1 g/ mL) = 45 g

Total mass of solution = (75 + 45) g = 120 g

Total heat gained by solution (Q) = mass of solution * specific heat capacity of solution * temperature change = 120 g * (4.184 J/ g ⁰C) * ( 24.1 - 20.3) ⁰ C = 1908 Joules

Moles of HCHO2 = volume in L * Molarity = (75/ 1000) L * 1.07 M = 0.08025 mole

Heat of reaction per mole of acid = 1908 J/ 0.08025 mole = 23776 J / mole = 23.776 kJ/ mole