A calorimeter contains $70.2 g$ of water at ${15.3}^{o} C$ . If $143.7 g$ of water at ${36.5}^{o} C$ in mixed it with the common temperature is ${28.7}^{o} C$ . The water equivalent of the calorimeter is:

15.6 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9.4 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.3 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

13.4 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $13.4 g$
Assume water equilient is $= n g m$
So, $(70.2) \times 1 \times (28.7 - 15.3) + n (28.7 - 15.3) = 143.7 \times 1 (36.5 - 28.7)$
$n (13.4) = 180.18$
$n = 13.4 g$

Suggest Corrections

Similar questions

A calorimeter contains 50 g of water at 50^0 C. The temperature falls to $45^{0}$ C in 10 minutes. When the calorimeter contains 100 g of water at $50^{0}$ C, it takes 18 minutes for the temperature to become $45^{0}$ C. Find the water equivalent of the calorimeter.