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# tricky af e=mcT (calorimetry) question watch

1. 30g of ice at 0.00 degrees Celcius is quickly added to a calorimeter at 24.20 degrees celcius. the final temperature comes to 2.60 degrees celcius. What is the heat capacity of the calorimeter. GIven fusion of ice is 334J/g and capacity of ice is 2.06 J/gC and cpacity of water is 4.18J/gC

i got 19.9 idk how can someone tell me what is happenung and how to solve it
2. (Original post by lolatmylifeee)
30g of ice at 0.00 degrees Celcius is quickly added to a calorimeter at 24.20 degrees celcius. the final temperature comes to 2.60 degrees celcius. What is the heat capacity of the calorimeter. GIven fusion of ice is 334J/g and capacity of ice is 2.06 J/gC and cpacity of water is 4.18J/gC

i got 19.9 idk how can someone tell me what is happenung and how to solve it
Do you know the mass of the calorimeter because if so that makes it easier.

The energy used in melting the ice E=Lm
Energy used heating the water to 2.6 degree Q=mcT (remember c is for water)
Add them together and you have the energy used there. That’s equal to the energy from cooling the calorimeter so if you have the mass you could rearrange and solve as needed.

Afraid I can’t help otherwise.
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