Question

Question: How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added?

Short answer

The rise in the temperature of water = 31.72\(^0C\)

Step-by-step solution

Specific heat

The heat required to raise the temperature of a substance is given by the formula Q = C × m ×∆ T.

Where “C” is the specific heat of the substance:

“m” is the mass of the substance, and

“∆T” is the change in the temperature of the substance.

Increase in temperature

We know from the given details that:

C = 4.184 J/g °C(Table 5.1)

m = 275 g

Q = 36.5kJ = 36500 J

By putting the values above in the equation Q = C × m × ∆ T, we get:

36500 = 4.184\( \times \)275\( \times \)\(\Delta \)T.

\(\Delta \)T = \(\frac{{36500}}{{4.184 \times 275}} = {31.72^0}C\).

Therefore, the rise in the temperature of 275 g of water = 31.72\(^0C\).