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# How many calories are needed to heat 100 g of water from 278 K to 288 K ?a.) 10 calories b.) 60 caloriesc.) 100 caloriesd.) 1000 calories

Last updated date: 17th Sep 2024
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Hint: Energy is the ability to do work. It is expressed in Joules, calories etc. In this question, we are asked the amount of energy required in calories to heat a specific amount of water to 10 K rise in temperature. This can be found by formula as -
Q = nCΔT
Where n = mass given
C = specific heat capacity of water
ΔT = change in temperature given.

Let us first write the things given to us and what we need to find out.
Given :
Mass of water = 100 g
Initial temperature = 278 K
Final temperature = 288 K

To find :
Number of calories required to heat this amount of water
We can find out the change in temperature as -
ΔT = Final temperature - Initial temperature
ΔT = (288 K - 278 K)
ΔT = 10 K
We have the formula for finding the energy as -
Q = nCΔT
Where n = mass given
C = specific heat capacity of water
ΔT = change in temperature given.
From question, we have n = 100 g
ΔT = 10 K
Further, C = 1 J -$^0{C^{ - 1}}{g^{ - 1}}$
So, putting all the values; we have
Q = 100 $\times$ 1$\times$10
Q = 1000 calories
Thus, 1000 calories are required to heat 100 g of the mass of water from 278 K to 288 K.
So, the correct answer is “Option D”.

Note: It must be noted that 1 calorie is equal to 4.184 Joules. And 1 K Calorie is equal to 1000 calories. So, we can also say that we require 1 K calorie amount of energy but we are given options in calories. So, we will move according to that.