1000 calories of heat are added to 100 gm of water at temperature 50 deg centigrade the resultant temperature of water will be?

a.10degC

b.30degC

c.60degC

d.80degC

To find the resultant temperature of the water after adding 1000 calories of heat, we can use the specific heat formula. The specific heat of water is approximately 1 calorie per gram per degree Celsius. This means that it takes 1 calorie to raise the temperature of 1 gram of water by 1 degree Celsius.

Calculating the Temperature Change

We start with the formula:

Q = mcΔT

• Q = heat added (in calories)
• m = mass of the water (in grams)
• c = specific heat capacity (for water, it's 1 cal/g°C)
• ΔT = change in temperature (in °C)

In this case:

• Q = 1000 calories
• m = 100 grams
• c = 1 cal/g°C

Finding the Change in Temperature

We can rearrange the formula to solve for ΔT:

ΔT = Q / (mc)

Substituting the values we have:

ΔT = 1000 calories / (100 grams × 1 cal/g°C)

Calculating this gives:

ΔT = 1000 / 100 = 10°C

Determining the Final Temperature

Now that we know the change in temperature (ΔT), we can find the final temperature of the water. The initial temperature of the water is 50°C. Therefore, we add the change in temperature to the initial temperature:

Final Temperature = Initial Temperature + ΔT

Final Temperature = 50°C + 10°C = 60°C

Conclusion

The resultant temperature of the water after adding 1000 calories of heat will be 60°C. Therefore, the correct answer is option c. 60°C.