Question

How many calories of are required to heat 731 grams of water from 35°c to 83°c? (assume that the specific heat of water is 1.00 cal/g·°c)?

Answer 1

Q=McDeltaT
Q=731g×1.00cal/g°c×(83°c-35°c)
Q=35088cal
therefore, their are 35088 calories are required to heat up 731g of water.

Answer 2

The calories required = 35088

Step-by-step explanation:

The heat required can be calculated by using specific heat of the water and mass and temperature raised.

The specific heat of water is the amount of heat required to raise the temperature of one gram of water by one degree celsius

Thus

Heat required = mass of water X specific heat X change in temperature

Given:

Specific heat of water = 1 cal /g °C

mass of water = 731 g

change in temperature = 83-35 = 48°C

Putting values

`Q=731X1X48=35088cal=35.088kcal`