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 data: 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