Question

Knowing that its specific heat capacity is 4.18 J/g °C , how much would the temperature (in
degrees C) of 16.0 g of pure water increase by when 670.J of heat are applied?

Answer 1

Given the following data:

• Specific heat capacity = 4.18 J/g °C.

• Mass of pure water = 16.0 grams.

• Quantity of heat = 670 Joules.

To calculate the change in temperature:

How to calculate change in temperature.

Mathematically, quantity of heat is given by this formula:

$Q = mc \theta$

Where:

• m is the mass.

• c is the specific heat capacity.

• $\theta$ is the change in temperature.

Making $\theta$ the subject of formula, we have:

$\theta = \frac{Q}{mc}$

Substituting the given parameters into the formula, we have:

$\theta = \frac{670}{16 \times 4.18} \\\\\theta = \frac{670}{66.88}$

Change in temperature = 10.02°C.

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