What is the amount of heat absorbed

when the temperature of 75 grams

of water increases from

20° C to 35°C

1125 cal

4707 Joules

Explanation:

This question can be solved by the calorimetry formula:

Q = m. C. ΔT

ΔT = Final T° - Initial T°

C = Specific heat of the compound

In this case our compound is water; the specific heat is 1 cal/g°C

m = mass

We replace dа ta: Q = 75 g. 1 cal/g°C (35°C - 20°C)

Q = 1125 cal

We can give this value in units of Joules

Let's do the conversion: 1 cal = 4.184 J

1125 cal. 4.184 J / 1 cal = 4707 Joules