Question

How much heat is released when 30 g of water at 96°C cools to 25°C? The specific heat of water
is 1 cal g-1°C-1.​

Answer 1

Answer:

2130 cal

Explanation:

H=mcΔT

C=Specific heat=1cal/gC

m=30grams

ΔT=96-25=71

H=30*1*71

  =2130cal

Answer 2

Given - Mass and specific heat

Find - Heat released

Solution - Heat released = mass*specific heat of water*change in temperature

Change in temperature = new temperature - original temperature

Change in temperature = 25 - 96

Performing subtraction

Change in temperature = -71° C

Keep the values in formula to find heat released.

Heat released = 30*1*(-71)

Performing multiplication

Heat released = -2130 Cal

Hence, the heat released is -2130 calories.