These are on specific heat

Question

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1. Calculate the energy required to heat a beaker of water at 23°C to boiling. The mass of the water is 68.9 g.
2. Determine the temperature change that will occur when 269 J of energy is applied to 24.7 g of gold.
3. A silver ring has a mass of 137.62 g. How many calories of heat are required to increase the temperature from 13.4°C to 173.6°C?
4. What is the mass of copper that increases its temperature by 312°C when 186,000 J of energy is applied?
5. A 278.96 g ingot of unknown metal is heated from 22.89°C to 267.29°C. This requires 3.91 kcal of energy. Calculate the specific heat of the metal and determine its identity.

Here's the table to help you find the answer.

Substance

Specific Heat Capacity

(J/g°C)

Water:4.184 (J/g°C)

Aluminum:0.89 (J/g°C)

Silicon:0.703 (J/g°C)

Iron:0.45 (J/g°C)

Copper:0.387 (J/g°C)

Silver:0.24 (J/g°C)

Gold:0.129 (J/g°C)

Lead:0.128 (J/g°C)

Answer 1

The questions all deal with the equation q=mc(delta)T.

The q is the heat energy measured in joules. Be careful some of the questions have or want other units.

The m is the mass in grams. The c is the specific heat from the table.

The deltaT is the change in temperature Tfinal-Tinitial. The temperature should be in celsius.

For each questions write down the variables and fill them in. Some are hidden like the boiling point of water which is 100 degrees C. Then put them into the equation which you would solve for the missing variable.

For example

q = ? q = (68.9g)(4.184 J/gC)(100 C - 23 C)

m = 68.9 22197.38J

c= 4.184 J/gC

Tfinal = 23 C

Tinitial =100 C

Try it.

Answer 2

When there is straight heating without a phase change, use Q = mC_pΔT as the equation where Q is the heat energy (most of these problems use Joules but it's often Calories or KCalories), "m" is the mass of the substance being heated (often times in grams), Cp is the Specific Heat of that material (often times in J/(g°C), and ΔT is the temperature change, T₂ - T₁ or temperature at time 2 minus temperature at time 1 (often times in °C).

Problem 1)

Q = mC_pΔT

Q = (68.9 g)(4.184 J/g°C)(100 °C - 23 °C)

Q = 22197 J but rounding to 1 sig fig yields:

Q = 22000 J

Problem 2)

Q = mC_pΔT

ΔT = Q/(mC_p)

ΔT = (269 J)/[(24.7 g)(0.129 J/g°C)

ΔT = 84.424 °C but rounding to 3 sig figs yields:

ΔT = 84.4 °C

The other problems are just like this except problem 5 requires conversion from kcal to joules before plugging in the energy number using 1 kcal = 4184 joules so 3.91 kcal is 16359.44 J.