Question

Determine the amount of heat energy in joules required to raise the temperature of 7.40g of water from 29.0OC to 46.0 OC.

Answer 1

Answer: The amount of heat energy in joules required to raise the temperature is 526 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed = ?

m= mass of substance = 7.40 g

c = specific heat capacity = $4.184J/g^0C$

Initial temperature of the water = $T_i$ = 29.0°C

Final temperature of the water = $T_f$  = 46.0°C

Change in temperature ,$\Delta T=T_f-T_i=(46.0-29.0)^0C=17.0^0C$

Putting in the values, we get:

$Q=7.40g\times 4.184J/g^0C\times 17.0^0C$

$Q=526J$

The amount of heat energy in joules required to raise the temperature is 526 Joules

Answer 2

The amount of heat energy will be "526 J".

Given:

Mass,

• m = 7.40 g

Initial temperature,

• $T_i$ = 29.0°C

Final temperature,

• $T_f$ = 46.0°C

Change in temperature,

• $\Delta T = T_f -T_i$

              $= 46.0-29.0$

              $= 17.0^{\circ} C$

The amount of heat energy,

→ $Q = mc \Delta T$

By substituting the values, we get

      $= 7.40\times 4.184\times 17.0$

      $= 526 \ J$

Thus the above answer is right.

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