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Question

Question: How many calories are needed to heat 100 g of water from 278 K to 288 K ? a.) 10 calories b.) 6...

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

Explanation

Solution

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.

Complete step by step answer:
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 -0C1g1^0{C^{ - 1}}{g^{ - 1}}
So, putting all the values; we have
Q = 100 ×\times 1×\times10
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.